local mma = math.max
local mfl, mce, mmi = math.floor, math.ceil, math.min

local AvlTree = {}
AvlTree.makenode = function(self, val, parent)
  local i = self.box[#self.box]
  if not i then i = #self.v + 1
  else table.remove(self.box)
  end
  self.v[i], self.p[i] = val, parent
  self.lc[i], self.rc[i], self.l[i], self.r[i] = 0, 0, 1, 1
  return i
end
AvlTree.create = function(self, lessthan, n)
  self.lessthan = lessthan
  self.root = 1
  self.box = {}
  for i = n + 1, 2, -1 do table.insert(self.box, i) end
  -- value, leftCount, rightCount, left, right, parent
  self.v, self.lc, self.rc, self.l, self.r, self.p = {}, {}, {}, {}, {}, {}
  for i = 1, n + 1 do
    self.v[i], self.p[i] = 0, 1
    self.lc[i], self.rc[i], self.l[i], self.r[i] = 0, 0, 1, 1
  end
end

AvlTree.recalc = function(self, i)
  local kl, kr = self.l[i], self.r[i]
  if 1 < kl then self.lc[i] = 1 + mma(self.lc[kl], self.rc[kl])
  else self.lc[i] = 0
  end
  if 1 < kr then self.rc[i] = 1 + mma(self.lc[kr], self.rc[kr])
  else self.rc[i] = 0
  end
end
AvlTree.recalcAll = function(self, i)
  while 1 < i do
    self:recalc(i)
    i = self.p[i]
  end
end

AvlTree.rotR = function(self, parent)
  local granp, child = self.p[parent], self.l[parent]
  self.r[child], self.l[parent] = parent, self.r[child]
  self.p[child], self.p[parent] = granp, child
  self.p[self.l[parent]] = parent
  if 1 < granp then
    if self.l[granp] == parent then
      self.l[granp] = child
    else
      self.r[granp] = child
    end
  else
    self.root = child
  end
end

AvlTree.rotL = function(self, parent)
  local granp, child = self.p[parent], self.r[parent]
  self.l[child], self.r[parent] = parent, self.l[child]
  self.p[child], self.p[parent] = granp, child
  self.p[self.r[parent]] = parent
  if 1 < granp then
    if self.r[granp] == parent then
      self.r[granp] = child
    else
      self.l[granp] = child
    end
  else
    self.root = child
  end
end

AvlTree.rotLR = function(self, lparent, rparent)
  local sp, sl, sr = self.p, self.l, self.r
  local granp, d = sp[rparent], sr[lparent]
  sp[lparent], sr[lparent] = d, sl[d]
  sp[rparent], sl[rparent] = d, sr[d]
  sp[sl[d]], sp[sr[d]] = lparent, rparent
  sp[d], sl[d], sr[d] = granp, lparent, rparent
  if 1 < granp then
    if sr[granp] == rparent then sr[granp] = d
    else sl[granp] = d
    end
  else self.root = d
  end
end

AvlTree.rotRL = function(self, rparent, lparent)
  local sp, sl, sr = self.p, self.l, self.r
  local granp, d = sp[lparent], sl[rparent]
  sp[rparent], sl[rparent] = d, sr[d]
  sp[lparent], sr[lparent] = d, sl[d]
  sp[sr[d]], sp[sl[d]] = rparent, lparent
  sp[d], sr[d], sl[d] = granp, rparent, lparent
  if 1 < granp then
    if sl[granp] == lparent then sl[granp] = d
    else sr[granp] = d
    end
  else self.root = d
  end
end

AvlTree.push = function(self, val)
  if self.root <= 1 then self.root = self:makenode(val, 1) return end
  local pos = self.root
  while true do
    if self.lessthan(val, self.v[pos]) then
      if 1 < self.l[pos] then
        pos = self.l[pos]
      else
        self.l[pos] = self:makenode(val, pos)
        pos = self.l[pos]
        break
      end
    else
      if 1 < self.r[pos] then
        pos = self.r[pos]
      else
        self.r[pos] = self:makenode(val, pos)
        pos = self.r[pos]
        break
      end
    end
  end
  while 1 < pos do
    local child, parent = pos, self.p[pos]
    if parent <= 1 then
      break
    end
    self:recalc(parent)
    local lcp_m_rcp = self.lc[parent] - self.rc[parent]
    if lcp_m_rcp % 2 ~= 0 then -- 1 or -1
      pos = parent
    elseif lcp_m_rcp == 2 then
      if self.lc[child] - 1 == self.rc[child] then
        self:rotR(parent)
        self:recalcAll(parent)
      else
        self:rotLR(child, parent)
        self:recalc(child)
        self:recalcAll(parent)
      end
      break
    elseif lcp_m_rcp == -2 then
      if self.rc[child] - 1 == self.lc[child] then
        self:rotL(parent)
        self:recalcAll(parent)
      else
        self:rotRL(child, parent)
        self:recalc(child)
        self:recalcAll(parent)
      end
      break
    else
      break
    end
  end
end

AvlTree.rmsub = function(self, node)
  while 1 < node do
    self:recalc(node)
    if self.lc[node] == self.rc[node] then
      node = self.p[node]
    elseif self.lc[node] + 1 == self.rc[node] then
      self:recalcAll(self.p[node])
      break
    else
      if self.lc[self.r[node]] == self.rc[self.r[node]] then
        self:rotL(node)
        self:recalcAll(node)
        break
      elseif self.lc[self.r[node]] + 1 == self.rc[self.r[node]] then
        local nr = self.r[node]
        self:rotL(node)
        self:recalc(node)
        node = nr
      else
        local nr = self.r[node]
        local nrl = self.l[nr]
        self:rotRL(nr, node)
        self:recalc(nr)
        self:recalc(node)
        node = nrl
      end
    end
  end
end

AvlTree.pop = function(self)
  local node = self.root
  while 1 < self.l[node] do
    node = self.l[node]
  end
  local v = self.v[node]
  local kp = self.p[node]
  self.p[self.r[node]] = kp
  if 1 < kp then
    self.l[kp] = self.r[node]
    self:rmsub(kp)
  else
    self.root = self.r[node]
  end
  table.insert(self.box, node)
  return v
end

AvlTree.top = function(self)
  local node = self.root
  while 1 < self.l[node] do
    node = self.l[node]
  end
  return self.v[node]
end

AvlTree.new = function(lessthan, n)
  local obj = {}
  setmetatable(obj, {__index = AvlTree})
  obj:create(lessthan, n)
  return obj
end

local n = io.read("*n")
local a = {}
for i = 1, n do
  a[i] = io.read("*n")
end
if n == 1 then print("YES") os.exit() end
local b = {}
for i = 1, n - 1 do
  b[i] = a[i + 1] - a[i]
end
b[n] = a[1] - a[n]

local avl = AvlTree.new(function(x, y) return b[x] < b[y] end, n)
local cnt = 0
local asked = {}
for i = 1, n do
  asked[i] = false
  avl:push(i)
end
while true do
  local p = avl:pop()
  if asked[p] then break end
  asked[p] = true
  local next_top = avl:top()
  if b[p] < cnt then
    local addcnt = mce((b[next_top] - b[p]) / n)
    addcnt = mma(addcnt, 1)
    b[p] = b[p] + addcnt * n
    cnt = cnt + addcnt
    avl:push(p)
  else
    break
  end
end
local valid = true
for i = 1, n do
  if cnt ~= b[i] then
    valid = false
    break
  end
end
print(valid and "YES" or "NO")

Score : 500 points

Problem Statement

There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.

Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.

Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO


local n = io.read("*n")
local t = {}
local sum = 0
for i = 1, n do
  t[i] = io.read("*n")
  sum = sum + t[i]
end
if (sum * 2) % (n * (n + 1)) ~= 0 then
  print("NO")
else
  local mul = (sum * 2) // (n * (n + 1))
  local isok = true
  local c = 0
  for i = 1, n do
    local sub = i == 1 and t[1] - t[n] or t[i] - t[i - 1]
    if (mul - sub) % n == 0 and 0 <= mul - sub then
      c = c + (mul - sub) // n
    else
      isok = false
      break
    end
  end
  isok = isok and c == mul
  print(isok and "YES" or "NO")
end

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


local mmi = math.min
local n = io.read("*n")
local c_term, c_thru = {}, {}
local edge = {}
local parent = {}
local childcnt = {}
for i = 1, n do
  c_term[i] = 0
  c_thru[i] = io.read("*n")
  edge[i] = {}
  parent[i] = -1
  childcnt[i] = 0
end
parent[1] = 0
for i = 1, n - 1 do
  local a, b = io.read("*n", "*n")
  table.insert(edge[a], b)
  table.insert(edge[b], a)
end
local tasks = {1}
for i = 1, n do
  local src = tasks[i]
  for j = 1, #edge[src] do
    local dst = edge[src][j]
    if parent[dst] == -1 then
      parent[dst] = src
      childcnt[src] = childcnt[src] + 1
      table.insert(tasks, dst)
    end
  end
end
tasks = {}
for i = 1, n do
  if childcnt[i] == 0 then
    c_term[i], c_thru[i] = c_thru[i], 0
    table.insert(tasks, i)
  end
end
local valid = true
for i = 1, n - 1 do
  local c = tasks[i]
  local p = parent[c]
  if c_term[p] + c_thru[p] < c_term[c] then
    valid = false
    break
  end
  local sub = mmi(c_term[c], c_thru[p])
  c_term[c] = c_term[c] - sub
  c_thru[p] = c_thru[p] - sub
  c_term[p] = c_term[p] + sub - c_term[c]
  childcnt[p] = childcnt[p] - 1
  if 0 == childcnt[p] then
    if 0 < c_thru[p] then
      valid = false
      break
    end
    table.insert(tasks, p)
  end
end
print(valid and "YES" or "NO")

Score : 700 points

Problem Statement

There is a tree with N vertices, numbered 1 through N.
The i-th of the N-1 edges connects vertices a_i and b_i.

Currently, there are A_i stones placed on vertex i.
Determine whether it is possible to remove all the stones from the vertices by repeatedly performing the following operation:

Select a pair of different leaves. Then, remove exactly one stone from every vertex on the path between those two vertices.
Here, a leaf is a vertex of the tree whose degree is 1, and the selected leaves themselves are also considered as vertices on the path connecting them.

Note that the operation cannot be performed if there is a vertex with no stone on the path.

Constraints

2 ≦ N ≦ 10^5

1 ≦ a_i,b_i ≦ N

0 ≦ A_i ≦ 10^9

The given graph is a tree.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

If it is possible to remove all the stones from the vertices, print YES. Otherwise, print NO.

Sample Input 1

5
1 2 1 1 2
2 4
5 2
3 2
1 3

Sample Output 1

YES

All the stones can be removed, as follows:

Select vertices 4 and 5. Then, there is one stone remaining on each vertex except 4.

Select vertices 1 and 5. Then, there is no stone on any vertex.

Sample Input 2

3
1 2 1
1 2
2 3

Sample Output 2

NO

Sample Input 3

6
3 2 2 2 2 2
1 2
2 3
1 4
1 5
4 6

Sample Output 3

YES


print(io.read("*n") < 1200 and "ABC" or "ARC")

Score : 100 points

Problem Statement

Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise.

You are given Smeke's current rating, x. Print ABC if Smeke will participate in ABC, and print ARC otherwise.

Constraints

1 ≦ x ≦ 3{,}000

x is an integer.

Input

The input is given from Standard Input in the following format:

x

Output

Print the answer.

Sample Input 1

1000

Sample Output 1

ABC

Smeke's current rating is less than 1200, thus the output should be ABC.

Sample Input 2

2000

Sample Output 2

ARC

Smeke's current rating is not less than 1200, thus the output should be ARC.


local x = io.read("n")
local r = x % 11
if r == 0 then
    print((x//11) * 2)
else
    if r <= 6 then
        print((x//11)*2 + 1)
    else
        print((x//11)*2 + 2)
    end
end

Score : 300 points

Problem Statement

Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7.

Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation:

Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward.

For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure.
If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward.
Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back.

Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.

Constraints

1 ≦ x ≦ 10^{15}

x is an integer.

Input

The input is given from Standard Input in the following format:

x

Output

Print the answer.

Sample Input 1

7

Sample Output 1

2

Sample Input 2

149696127901

Sample Output 2

27217477801


local n = io.read("*n")
local t = {}
for i = 1, n do
  local a = io.read("*n")
  if not t[a] then t[a] = 0 else t[a] = t[a] + 1 end
end
local rm = 0
for k, v in pairs(t) do
  rm = rm + v
end
if rm % 2 == 1 then rm = rm + 1 end
print(n - rm)

Score : 400 points

Problem Statement

Snuke has decided to play a game using cards.
He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.

He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.

Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.

Constraints

3 ≦ N ≦ 10^{5}

N is odd.

1 ≦ A_i ≦ 10^{5}

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_{N}

Output

Print the answer.

Sample Input 1

5
1 2 1 3 7

Sample Output 1

3

One optimal solution is to perform the operation once, taking out two cards with 1 and one card with 2. One card with 1 and another with 2 will be eaten, and the remaining card with 1 will be returned to deck. Then, the values written on the remaining cards in the deck will be pairwise distinct: 1, 3 and 7.

Sample Input 2

15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1

Sample Output 2

7


local n = io.read("*n")
local cnt = n // 11
n, cnt = n - cnt * 11, cnt * 2
if(1 <= n) then
  cnt = cnt + (n <= 6 and 1 or 2)
end
print(cnt)


local n = io.read("*n")
local s = {}
local ans = 0
for _ = 1,n do
    local a = io.read("*n")
    if s[a] == nil then
        s[a] = true
        ans = ans + 1
    end
end
if ans % 2 == 0 then
    ans = ans - 1
end
print(ans)


local mfl = math.floor
local n, m = io.read("*n", "*n")
local imos = {}
for i = 1, m + 1 do
  imos[i] = 0
end
for i_n = 1, n do
  local l, r = io.read("*n", "*n")
  l = l - 1
  imos[1] = imos[1] + 1
  imos[r + 1] = imos[r + 1] - 1
  for i = 1, r do
    local rlim2 = mfl(r / (i + 1))
    local llim = mfl(l / i)
    if rlim2 < llim then
      imos[rlim2 + 1] = imos[rlim2 + 1] - 1
      imos[llim + 1] = imos[llim + 1] + 1
    else
      break
    end
  end
end
for i = 2, m do
  imos[i] = imos[i] + imos[i - 1]
end
imos[m + 1] = nil
print(table.concat(imos, "\n"))

Score : 700 points

Problem Statement

Snuke has decided to play a game, where the player runs a railway company.
There are M+1 stations on Snuke Line, numbered 0 through M.
A train on Snuke Line stops at station 0 and every d-th station thereafter, where d is a predetermined constant for each train.
For example, if d = 3, the train stops at station 0, 3, 6, 9, and so forth.

There are N kinds of souvenirs sold in areas around Snuke Line. The i-th kind of souvenirs can be purchased when the train stops at one of the following stations: stations l_i, l_i+1, l_i+2, ..., r_i.

There are M values of d, the interval between two stops, for trains on Snuke Line: 1, 2, 3, ..., M.
For each of these M values, find the number of the kinds of souvenirs that can be purchased if one takes a train with that value of d at station 0.
Here, assume that it is not allowed to change trains.

Constraints

1 ≦ N ≦ 3 × 10^{5}

1 ≦ M ≦ 10^{5}

1 ≦ l_i ≦ r_i ≦ M

Input

The input is given from Standard Input in the following format:

N M
l_1 r_1
:
l_{N} r_{N}

Output

Print the answer in M lines. The i-th line should contain the maximum number of the kinds of souvenirs that can be purchased if one takes a train stopping every i-th station.

Sample Input 1

3 3
1 2
2 3
3 3

Sample Output 1

3
2
2

If one takes a train stopping every station, three kinds of souvenirs can be purchased: kind 1, 2 and 3.

If one takes a train stopping every second station, two kinds of souvenirs can be purchased: kind 1 and 2.

If one takes a train stopping every third station, two kinds of souvenirs can be purchased: kind 2 and 3.

Sample Input 2

7 9
1 7
5 9
5 7
5 9
1 1
6 8
3 4

Sample Output 2

7
6
6
5
4
5
5
3
2


local mfl, mce = math.floor, math.ceil
local n = io.read("*n")
local a, b = {}, {}
for i = 1, n do
  a[i], b[i] = io.read("*n", "*n")
end
local sum = 0
for i = n, 1, -1 do
  local dst = mce((a[i] + sum) / b[i])
  sum = sum + dst * b[i] - (a[i] + sum)
end
print(sum)

Score : 300 points

Problem Statement

There are an integer sequence A_1,...,A_N consisting of N terms, and N buttons.
When the i-th (1 ≦ i ≦ N) button is pressed, the values of the i terms from the first through the i-th are all incremented by 1.

There is also another integer sequence B_1,...,B_N. Takahashi will push the buttons some number of times so that for every i, A_i will be a multiple of B_i.

Find the minimum number of times Takahashi will press the buttons.

Constraints

All input values are integers.

1 ≦ N ≦ 10^5

0 ≦ A_i ≦ 10^9(1 ≦ i ≦ N)

1 ≦ B_i ≦ 10^9(1 ≦ i ≦ N)

Input

The input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Output

Print an integer representing the minimum number of times Takahashi will press the buttons.

Sample Input 1

3
3 5
2 7
9 4

Sample Output 1

7

Press the first button twice, the second button twice and the third button three times.

Sample Input 2

7
3 1
4 1
5 9
2 6
5 3
5 8
9 7

Sample Output 2

22


local n = io.read("*n")
local parent, child, childpush = {}, {}, {}
local depth = {}
for i = 1, n do
  child[i] = {}
  depth[i] = 0
  childpush[i] = 0
end
parent[1] = 0
for i = 2, n do
  local a = io.read("*n")
  parent[i] = a
  table.insert(child[a], i)
end
local tasks = {}
for i = 1, n do
  if #child[i] == 0 then
    table.insert(tasks, i)
  end
end
local function sortfunc(x, y)
  return depth[x] > depth[y]
end
local mma = math.max
while 0 < #tasks do
  local c = tasks[#tasks]
  table.remove(tasks)
  local p = parent[c]
  childpush[p] = childpush[p] + 1
  if childpush[p] == #child[p] then
    table.sort(child[p], sortfunc)
    local d = 0
    for i = 1, #child[p] do
      d = mma(d, i + depth[child[p][i]])
    end
    depth[p] = d
    if 1 < p then
      table.insert(tasks, p)
    end
  end
end
print(depth[1])

Score : 800 points

Problem Statement

N contestants participated in a competition. The total of N-1 matches were played in a knockout tournament.
For some reasons, the tournament may not be "fair" for all the contestants.
That is, the number of the matches that must be played in order to win the championship may be different for each contestant. The structure of the tournament is formally described at the end of this statement.

After each match, there were always one winner and one loser. The last contestant standing was declared the champion.

Figure: an example of a tournament

For convenience, the contestants were numbered 1 through N. The contestant numbered 1 was the champion, and the contestant numbered i(2 ≦ i ≦ N) was defeated in a match against the contestant numbered a_i.

We will define the depth of the tournament as the maximum number of the matches that must be played in order to win the championship over all the contestants.

Find the minimum possible depth of the tournament.

The formal description of the structure of the tournament is as follows. In the i-th match, one of the following played against each other:

Two predetermined contestants

One predetermined contestant and the winner of the j-th match, where j(j<i) was predetermined

The winner of the j-th match and the winner of the k-th match, where j and k (j,k<i, j ≠ k) were predetermined

Such structure is valid structure of the tournament, if and only if no contestant who has already been defeated in a match will never play in a match, regardless of the outcome of the matches.

Constraints

2 ≦ N ≦ 10^5

1 ≦ a_i ≦ N(2 ≦ i ≦ N)

It is guaranteed that the input is consistent (that is, there exists at least one tournament that matches the given information).

Input

The input is given from Standard Input in the following format:

N
a_2
:
a_N

Output

Print the minimum possible depth of the tournament.

Sample Input 1

5
1
1
2
4

Sample Output 1

3

The following tournament and the result of the matches are consistent with the given information:

In the first match, contestants 4 and 5 played against each other, and contestant 4 won.

In the second match, contestants 2 and 4 played against each other, and contestant 2 won.

In the third match, contestants 1 and 3 played against each other, and contestant 1 won.

In the fourth match, contestants 1 and 2 played against each other, and contestant 1 won.

The depth of this tournament is 3, since contestant 5 must play three matches in order to win the championship.
There is no tournament with depth 2 or less that is consistent with the given information, thus the output should be 3.

Sample Input 2

7
1
2
1
3
1
4

Sample Output 2

3

Sample Input 3

4
4
4
1

Sample Output 3

3


Score : 100 points

Problem Statement

There are two rectangles.
The lengths of the vertical sides of the first rectangle are A, and the lengths of the horizontal sides of the first rectangle are B.
The lengths of the vertical sides of the second rectangle are C, and the lengths of the horizontal sides of the second rectangle are D.

Print the area of the rectangle with the larger area.
If the two rectangles have equal areas, print that area.

Constraints

All input values are integers.

1≤A≤10^4

1≤B≤10^4

1≤C≤10^4

1≤D≤10^4

Input

The input is given from Standard Input in the following format:

A B C D

Output

Print the area of the rectangle with the larger area.
If the two rectangles have equal areas, print that area.

Sample Input 1

3 5 2 7

Sample Output 1

15

The first rectangle has an area of 3×5=15, and the second rectangle has an area of 2×7=14.
Thus, the output should be 15, the larger area.

Sample Input 2

100 600 200 300

Sample Output 2

60000


local mod = 1000000007
local function bmul(x, y) return (x * y) % mod end
local n = io.read("*n")
local ret = 1
local t = {}
for i = 1, n do t[i] = true end
for i = 2, n do
  if t[i] then
    local c = 0
    for j = i, n, i do
      t[j] = false
      while j % i == 0 do
        c = c + 1
        j = j // i
      end
    end
    ret = bmul(ret, (c + 1))
  end
end
print(ret)

Score : 300 points

Problem Statement

You are given an integer N.
Find the number of the positive divisors of N!, modulo 10^9+7.

Constraints

1≤N≤10^3

Input

The input is given from Standard Input in the following format:

N

Output

Print the number of the positive divisors of N!, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

4

There are four divisors of 3! =6: 1, 2, 3 and 6. Thus, the output should be 4.

Sample Input 2

6

Sample Output 2

30

Sample Input 3

1000

Sample Output 3

972926972


local n,a,b=io.read("n","n","n")
local x={}
for i=1,n do
    x[i]=io.read("n")
end

local cost=0
for i=1,n-1 do
    cost=cost+math.min(a*(x[i+1]-x[i]),b)
end
print(cost)

Score : 500 points

Problem Statement

There are N towns on a line running east-west.
The towns are numbered 1 through N, in order from west to east.
Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value.
The coordinate of town i is X_i.

You are now at town 1, and you want to visit all the other towns.
You have two ways to travel:

Walk on the line.
Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.

Teleport to any location of your choice.
Your fatigue level increases by B, regardless of the distance covered.

Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.

Constraints

All input values are integers.

2≤N≤10^5

1≤X_i≤10^9

For all i(1≤i≤N-1), X_i<X_{i+1}.

1≤A≤10^9

1≤B≤10^9

Input

The input is given from Standard Input in the following format:

N A B
X_1 X_2 ... X_N

Output

Print the minimum possible total increase of your fatigue level when you visit all the towns.

Sample Input 1

4 2 5
1 2 5 7

Sample Output 1

11

From town 1, walk a distance of 1 to town 2, then teleport to town 3, then walk a distance of 2 to town 4.
The total increase of your fatigue level in this case is 2×1+5+2×2=11, which is the minimum possible value.

Sample Input 2

7 1 100
40 43 45 105 108 115 124

Sample Output 2

84

From town 1, walk all the way to town 7.
The total increase of your fatigue level in this case is 84, which is the minimum possible value.

Sample Input 3

7 1 2
24 35 40 68 72 99 103

Sample Output 3

12

Visit all the towns in any order by teleporting six times.
The total increase of your fatigue level in this case is 12, which is the minimum possible value.


local mma, mmi = math.max, math.min
local mce, mfl = math.ceil, math.floor

local mod = 1000000007
local function bmul(x, y)
  local x0, y0 = x % 31623, y % 31623
  local x1, y1 = mfl(x / 31623), mfl(y / 31623)
  return (x1 * y1 * 14122 + (x1 * y0 + x0 * y1) * 31623 + x0 * y0) % mod
end
local function badd(x, y)
  return (x + y) % mod
end

local function modpow(src, pow)
  local res = 1
  while 0 < pow do
    if pow % 2 == 1 then
      res = bmul(res, src)
      pow = pow - 1
    end
    src = bmul(src, src)
    pow = mfl(pow / 2)
  end
  return res
end

local function modinv(src)
  return modpow(src, mod - 2)
end

local fact = {1}
for i = 2, 1000 do
  fact[i] = bmul(fact[i - 1], i)
end
local factinv = {}
for i = 1, 1000 do
  factinv[i] = modinv(fact[i])
end

local function getComb(n, k)
  if n == k then return 1 end
  return bmul(fact[n], bmul(factinv[k], factinv[n - k]))
end

local n, a, b, c, d = io.read("*n", "*n", "*n", "*n", "*n")
local rem = {}
local remalt = {}
for i = 1, n do
  rem[i] = 0
  remalt[i] = 0
end
rem[n] = 1

local function calc(ia, ic)
  local ret = 1
  for i = 0, ic - 1 do
    ret = bmul(ret, getComb((ic - i) * ia, ia))
  end
  return bmul(ret, factinv[ic])
end

local ans = 0
for ia = a, b do
  local src = (ia - a) % 2 == 0 and rem or remalt
  local dst = (ia - a) % 2 == 0 and remalt or rem
  for i = 1, n do dst[i] = src[i] end
  local cmin = c
  local cmax = mmi(d, mfl(n / ia))
  for ic = cmin, cmax do
    local v = ic * ia
    local z = calc(ia, ic)
    ans = badd(ans, bmul(src[v], z))
    for i_n = v + 1, n do
      dst[i_n - v] = badd(dst[i_n - v], bmul(src[i_n], bmul(z, getComb(i_n, i_n - v))))
    end
  end
end
print(ans)

Score : 600 points

Problem Statement

There are N people, conveniently numbered 1 through N.
We want to divide them into some number of groups, under the following two conditions:

Every group contains between A and B people, inclusive.

Let F_i be the number of the groups containing exactly i people. Then, for all i, either F_i=0 or C≤F_i≤D holds.

Find the number of these ways to divide the people into groups.
Here, two ways to divide them into groups is considered different if and only if there exists two people such that they belong to the same group in exactly one of the two ways.
Since the number of these ways can be extremely large, print the count modulo 10^9+7.

Constraints

1≤N≤10^3

1≤A≤B≤N

1≤C≤D≤N

Input

The input is given from Standard Input in the following format:

N A B C D

Output

Print the number of ways to divide the people into groups under the conditions, modulo 10^9+7.

Sample Input 1

3 1 3 1 2

Sample Output 1

4

There are four ways to divide the people:

(1,2),(3)

(1,3),(2)

(2,3),(1)

(1,2,3)

The following way to divide the people does not count: (1),(2),(3). This is because it only satisfies the first condition and not the second.

Sample Input 2

7 2 3 1 3

Sample Output 2

105

The only ways to divide the people under the conditions are the ones where there are two groups of two people, and one group of three people.
There are 105 such ways.

Sample Input 3

1000 1 1000 1 1000

Sample Output 3

465231251

Sample Input 4

10 3 4 2 5

Sample Output 4

0

The answer can be 0.


s=io.read()
x = string.gsub(s,","," ")
print(x)

Score : 100 points

Problem Statement

As a New Year's gift, Dolphin received a string s of length 19.

The string s has the following format: [five lowercase English letters],[seven lowercase English letters],[five lowercase English letters].

Dolphin wants to convert the comma-separated string s into a space-separated string.

Write a program to perform the conversion for him.

Constraints

The length of s is 19.

The sixth and fourteenth characters in s are ,.

The other characters in s are lowercase English letters.

Input

The input is given from Standard Input in the following format:

s

Output

Print the string after the conversion.

Sample Input 1

happy,newyear,enjoy

Sample Output 1

happy newyear enjoy

Replace all the commas in happy,newyear,enjoy with spaces to obtain happy newyear enjoy.

Sample Input 2

haiku,atcoder,tasks

Sample Output 2

haiku atcoder tasks

Sample Input 3

abcde,fghihgf,edcba

Sample Output 3

abcde fghihgf edcba


-- heap
local Heap = {}
function Heap.new(upper_func)
    local self = setmetatable({}, {__index = Heap})
    self.a = {}
    self.upper_func = upper_func
    self.upper = function(k1, k2)
        assert(k1 < k2)
        if k1 == 0 then
            return false
        else
            return upper_func(self.a[k1], self.a[k2])
        end
    end
    self.lower = function(k1, k2)
        return not self.upper(k1, k2)
    end
    return self
end

function Heap:count()
    return #self.a
end

function Heap:_swap(k1, k2)
    local old_k1 = self.a[k1]
    self.a[k1] = self.a[k2]
    self.a[k2] = old_k1
end

function Heap:upheap(newk)
    local ku = newk // 2
    local k = newk
    while ku > 0 and self.lower(ku, k) do
        self:_swap(ku, k)
        k = k//2
        ku = ku//2
    end
end

function Heap:insertq(e)
    table.insert(self.a, e)
    self:upheap(#self.a)
end

function Heap:downheap(newk)
    local k = newk
    while k <= #self.a//2 do
        local kdl = k*2
        local kdr = kdl + 1
        local kd
        if kdr <= #self.a and self.lower(kdl, kdr) then
            kd = kdr
        else
            kd = kdl
        end
        if self.upper(k, kd) then
            break
        end
        self:_swap(k, kd)
        k = kd
    end
end

function Heap:top()
    assert(#self.a >= 1)
    return self.a[1]
end

function Heap:remove_top()
    assert(#self.a >= 1)
    local v = self.a[1]
    self:_swap(1, #self.a)
    table.remove(self.a)
    self:downheap(1)
    return v
end

function Heap:replace_top(e)
    assert(#self.a >= 1)
    local old_v = self.a[1]
    local swapped = false
    if self.upper_func(old_v, e) then
        self.a[1] = e
        self:downheap(1)
        swapped = true
    end
    return swapped, old_v
end

-- dijkstra's algorithm
local function find_shortest_path_all(adj_list, src)
    local INF = math.maxinteger
    local from1 = {}
    local prevs = {}
    from1[src] = 0
    local q = Heap.new(function (v1, v2)
        return from1[v1] < from1[v2]
    end)
    local proved = {}
    q:insertq(src)
    while q:count() > 0 do
        local cur_v = q:remove_top()
        if not proved[cur_v] then
            proved[cur_v] = true
            local cur_cost = from1[cur_v]
            --[[
            if cur_v == dst then
                return cur_cost
            end
            ]]
            for _, link in ipairs(adj_list[cur_v]) do
                local to, cost = table.unpack(link)
                if not proved[to] then
                    -- compare costs
                    local before = from1[to]
                    local candidate = cur_cost + cost
                    if not before or candidate < before then
                        from1[to] = candidate
                        prevs[to] = cur_v
                        q:insertq(to)
                    end
                end
            end
        end
    end
    return from1, prevs
end

local function make_path(prevs, src, dst)
    local path = {}
    path[#path+1] = dst
    local cur = dst
    while cur ~= src do
        local next = prevs[cur]
        path[#path+1] = next
        cur = next
    end
    -- reverse
    for i=1, #path//2 do
        path[i], path[#path-i+1] = path[#path-i+1], path[i]
    end
    return path
end

-- main
local N, M = io.read("n", "n")

local edges = {}
for i=1,M do
    local a, b, cost = io.read("n", "n", "n")
    edges[#edges+1] = {a, b, cost}
end

local adj_list = {}
for i=1,N do
    adj_list[i] = {}
end

local edge_check = {}
for s=1,N do
    edge_check[s] = {}
end

for i=1,M do
    local a, b, cost = table.unpack(edges[i])
    local list = adj_list[a]
    list[#list+1] = {b, cost}
    local list2 = adj_list[b]
    list2[#list2+1] = {a, cost}
    local small = math.min(a, b)
    local big = math.max(a, b)
    edge_check[small][big] = 0
end

for s=1,N do
    local c, prevs = find_shortest_path_all(adj_list, s)
    for dd=1,N do
        if dd ~= s then
            local ss = prevs[dd]
            local small = math.min(ss, dd)
            local big = math.max(ss, dd)
            edge_check[small][big] = edge_check[small][big] + 1
        end
    end
end

local ans = 0
for s=1,N do
    for d=s+1,N do
        local count = edge_check[s][d]
        if count and count == 0 then
            ans = ans + 1
        end
    end
end
print(ans)

Score : 400 points

Problem Statement

You are given an undirected connected weighted graph with N vertices and M edges that contains neither self-loops nor double edges.

The i-th (1≤i≤M) edge connects vertex a_i and vertex b_i with a distance of c_i.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

A connected graph is a graph where there is a path between every pair of different vertices.

Find the number of the edges that are not contained in any shortest path between any pair of different vertices.

Constraints

2≤N≤100

N-1≤M≤min(N(N-1)/2,1000)

1≤a_i,b_i≤N

1≤c_i≤1000

c_i is an integer.

The given graph contains neither self-loops nor double edges.

The given graph is connected.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the number of the edges in the graph that are not contained in any shortest path between any pair of different vertices.

Sample Input 1

3 3
1 2 1
1 3 1
2 3 3

Sample Output 1

1

In the given graph, the shortest paths between all pairs of different vertices are as follows:

The shortest path from vertex 1 to vertex 2 is: vertex 1 → vertex 2, with the length of 1.

The shortest path from vertex 1 to vertex 3 is: vertex 1 → vertex 3, with the length of 1.

The shortest path from vertex 2 to vertex 1 is: vertex 2 → vertex 1, with the length of 1.

The shortest path from vertex 2 to vertex 3 is: vertex 2 → vertex 1 → vertex 3, with the length of 2.

The shortest path from vertex 3 to vertex 1 is: vertex 3 → vertex 1, with the length of 1.

The shortest path from vertex 3 to vertex 2 is: vertex 3 → vertex 1 → vertex 2, with the length of 2.

Thus, the only edge that is not contained in any shortest path, is the edge of length 3 connecting vertex 2 and vertex 3, hence the output should be 1.

Sample Input 2

3 2
1 2 1
2 3 1

Sample Output 2

0

Every edge is contained in some shortest path between some pair of different vertices.


local n, k = io.read("*n", "*n")
local sum, psum = {}, {}
local cur, pcur = 0, 0
for i = 1, n do
  local a = io.read("*n")
  cur = cur + a
  sum[i] = cur
  if 0 < a then
    pcur = pcur + a
  end
  psum[i] = pcur
end
local mma = math.max
local ret = 0
for i = 1, n - k + 1 do
  local v = sum[i + k - 1]
  if 1 < i then
    v = v - sum[i - 1]
  end
  v = mma(0, v)
  v = v + psum[n] - psum[i + k - 1]
  if 1 < i then
    v = v + psum[i - 1]
  end
  ret = mma(ret, v)
end
print(ret)

Score : 400 points

Problem Statement

There are N squares aligned in a row.
The i-th square from the left contains an integer a_i.

Initially, all the squares are white.
Snuke will perform the following operation some number of times:

Select K consecutive squares. Then, paint all of them white, or paint all of them black. Here, the colors of the squares are overwritten.

After Snuke finishes performing the operation, the score will be calculated as the sum of the integers contained in the black squares.
Find the maximum possible score.

Constraints

1≤N≤10^5

1≤K≤N

a_i is an integer.

|a_i|≤10^9

Input

The input is given from Standard Input in the following format:

N K
a_1 a_2 ... a_N

Output

Print the maximum possible score.

Sample Input 1

5 3
-10 10 -10 10 -10

Sample Output 1

10

Paint the following squares black: the second, third and fourth squares from the left.

Sample Input 2

4 2
10 -10 -10 10

Sample Output 2

20

One possible way to obtain the maximum score is as follows:

Paint the following squares black: the first and second squares from the left.

Paint the following squares black: the third and fourth squares from the left.

Paint the following squares white: the second and third squares from the left.

Sample Input 3

1 1
-10

Sample Output 3

0

Sample Input 4

10 5
5 -4 -5 -8 -4 7 2 -4 0 7

Sample Output 4

17


Score : 600 points

Problem Statement

A tetromino is a figure formed by joining four squares edge to edge.
We will refer to the following seven kinds of tetromino as I-, O-, T-, J-, L-, S- and Z-tetrominos, respectively:

Snuke has many tetrominos. The number of I-, O-, T-, J-, L-, S- and Z-tetrominos in his possession are a_I, a_O, a_T, a_J, a_L, a_S and a_Z, respectively.
Snuke will join K of his tetrominos to form a rectangle that is two squares tall and 2K squares wide.
Here, the following rules must be followed:

When placing each tetromino, rotation is allowed, but reflection is not.

Each square in the rectangle must be covered by exactly one tetromino.

No part of each tetromino may be outside the rectangle.

Snuke wants to form as large a rectangle as possible.
Find the maximum possible value of K.

Constraints

0≤a_I,a_O,a_T,a_J,a_L,a_S,a_Z≤10^9

a_I+a_O+a_T+a_J+a_L+a_S+a_Z≥1

Input

The input is given from Standard Input in the following format:

a_I a_O a_T a_J a_L a_S a_Z

Output

Print the maximum possible value of K. If no rectangle can be formed, print 0.

Sample Input 1

2 1 1 0 0 0 0

Sample Output 1

3

One possible way to form the largest rectangle is shown in the following figure:

Sample Input 2

0 0 10 0 0 0 0

Sample Output 2

0

No rectangle can be formed.


local t = {}
for i = 1, 7 do t[i] = io.read("*n") end
local ret = t[2] + 2 * (t[1] // 2 + t[4] // 2 + t[5] // 2)
local a = t[1] % 2 + t[4] % 2 + t[5] % 2
if a == 3 then
  ret = ret + 3
elseif a == 2 then
  if 0 < t[1] and 0 < t[4] and 0 < t[5] then
    ret = ret + 1
  end
end
print(ret)


a,op,b = io.read("*n","*l","*n")
if string.find(op, "+") == nil then
	print(a - b)
else
	print(a + b)
end

Score : 100 points

Problem Statement

Joisino wants to evaluate the formula "A op B".
Here, A and B are integers, and the binary operator op is either + or -.
Your task is to evaluate the formula instead of her.

Constraints

1≦A,B≦10^9

op is either + or -.

Input

The input is given from Standard Input in the following format:

A op B

Output

Evaluate the formula and print the result.

Sample Input 1

1 + 2

Sample Output 1

3

Since 1 + 2 = 3, the output should be 3.

Sample Input 2

5 - 7

Sample Output 2

-2


t={}
t.a,t.i,t.u,t.e,t.o=0,0,0,0,0
s=io.read()
print(t[s] and "vowel" or "consonant")

Score : 100 points

Problem Statement

Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: a, e, i, o and u.

Constraints

c is a lowercase English letter.

Input

The input is given from Standard Input in the following format:

c

Output

If c is a vowel, print vowel. Otherwise, print consonant.

Sample Input 1

a

Sample Output 1

vowel

Since a is a vowel, print vowel.

Sample Input 2

z

Sample Output 2

consonant

Sample Input 3

s

Sample Output 3

consonant


a=io.read()
f=false
for i in {"a","e","i","o","u"}do
  f=f or (i==a)
end
print(f and "vowel"or"consonant")


h,w=io.read("*n","*n","*l")
for i=1,h do
    s=io.read()
    print(s)
    print(s)
end


local road={}
local rail={}
local size={}

local function init(n)
    for i=1,n do
        road[i]=i
        rail[i]=i
        size[i]=1
    end
end

local function root(x,list)
    return list[x]==x and x or root(list[x],list)
end

local function same(x,y,list)
    return root(x,list)==root(y,list)
end

local function unite(x,y,list)
    x=root(x,list)
    y=root(y,list)
    if x==y then
        return
    end
    if size[x]>size[y] then
        x,y=y,x
    end
    list[x]=y
    size[y]=size[y]+size[x]
end

local function getsize(x)
    return size[root(x,list)]
end

----------

local n,k,l=io.read("n","n","n")
init(n)

for i=1,k do
    local p,q=io.read("n","n")
    unite(p,q,road)
end

for i=1,l do
    local r,s=io.read("n","n")
    unite(r,s,rail)
end

local counter={}
for i=1,n do
    local key=root(i,road).." "..root(i,rail)
    counter[key]=(counter[key] or 0)+1
end

for i=1,n do
    local key=root(i,road).." "..root(i,rail)
    io.write(counter[key].." ")
end
print()

Score : 400 points

Problem Statement

There are N cities. There are also K roads and L railways, extending between the cities.
The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities.
No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.

We will say city A and B are connected by roads if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads.
We will also define connectivity by railways similarly.

For each city, find the number of the cities connected to that city by both roads and railways.

Constraints

2 ≦ N ≦ 2*10^5

1 ≦ K, L≦ 10^5

1 ≦ p_i, q_i, r_i, s_i ≦ N

p_i < q_i

r_i < s_i

When i ≠ j, (p_i, q_i) ≠ (p_j, q_j)

When i ≠ j, (r_i, s_i) ≠ (r_j, s_j)

Input

The input is given from Standard Input in the following format:

N K L
p_1 q_1
:
p_K q_K
r_1 s_1
:
r_L s_L

Output

Print N integers. The i-th of them should represent the number of the cities connected to the i-th city by both roads and railways.

Sample Input 1

4 3 1
1 2
2 3
3 4
2 3

Sample Output 1

1 2 2 1

All the four cities are connected to each other by roads.

By railways, only the second and third cities are connected. Thus, the answers for the cities are 1, 2, 2 and 1, respectively.

Sample Input 2

4 2 2
1 2
2 3
1 4
2 3

Sample Output 2

1 2 2 1

Sample Input 3

7 4 4
1 2
2 3
2 5
6 7
3 5
4 5
3 4
6 7

Sample Output 3

1 1 2 1 2 2 2


local n, k, l = io.read("*n", "*n", "*n")

local function uf_initialize(n)
  local parent = {}
  for i = 1, n do parent[i] = i end
  return parent
end

local function uf_findroot(idx, parent)
  local idx_update = idx
  while parent[idx] ~= idx do
    idx = parent[idx]
  end
  while parent[idx_update] ~= idx do
    parent[idx_update], idx_update = idx, parent[idx_update]
  end
  return idx
end

local p1, p2 = uf_initialize(n), uf_initialize(n)

for i = 1, k do
  local a, b = io.read("*n", "*n")
  local ra, rb = uf_findroot(a, p1), uf_findroot(b, p1)
  p1[b], p1[rb] = ra, ra
end

for i = 1, l do
  local a, b = io.read("*n", "*n")
  local ra, rb = uf_findroot(a, p2), uf_findroot(b, p2)
  p2[b], p2[rb] = ra, ra
end

local function getidx(r1, r2)
  return r1 - 1 + r2 * n
end
local groups = {}
for i = 1, n do
  local idx = getidx(uf_findroot(i, p1), uf_findroot(i, p2))
  if not groups[idx] then
    groups[idx] = 1
  else
    groups[idx] = groups[idx] + 1
  end
end
for i = 1, n do
  local idx = getidx(uf_findroot(i, p1), uf_findroot(i, p2))
  io.write(groups[idx])
  io.write(i == n and "\n" or " ")
end


Score : 100 points

Problem Statement

Snuke is going to open a contest named "AtCoder s Contest".
Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters.

Snuke has decided to abbreviate the name of the contest as "AxC".
Here, x is the uppercase English letter at the beginning of s.

Given the name of the contest, print the abbreviation of the name.

Constraints

The length of s is between 1 and 100, inclusive.

The first character in s is an uppercase English letter.

The second and subsequent characters in s are lowercase English letters.

Input

The input is given from Standard Input in the following format:

AtCoder s Contest

Output

Print the abbreviation of the name of the contest.

Sample Input 1

AtCoder Beginner Contest

Sample Output 1

ABC

The contest in which you are participating now.

Sample Input 2

AtCoder Snuke Contest

Sample Output 2

ASC

This contest does not actually exist.

Sample Input 3

AtCoder X Contest

Sample Output 3

AXC


Score : 200 points

Problem Statement

You are given nonnegative integers a and b (a ≤ b), and a positive integer x.
Among the integers between a and b, inclusive, how many are divisible by x?

Constraints

0 ≤ a ≤ b ≤ 10^{18}

1 ≤ x ≤ 10^{18}

Input

The input is given from Standard Input in the following format:

a b x

Output

Print the number of the integers between a and b, inclusive, that are divisible by x.

Sample Input 1

4 8 2

Sample Output 1

3

There are three integers between 4 and 8, inclusive, that are divisible by 2: 4, 6 and 8.

Sample Input 2

0 5 1

Sample Output 2

6

There are six integers between 0 and 5, inclusive, that are divisible by 1: 0, 1, 2, 3, 4 and 5.

Sample Input 3

9 9 2

Sample Output 3

0

There are no integer between 9 and 9, inclusive, that is divisible by 2.

Sample Input 4

1 1000000000000000000 3

Sample Output 4

333333333333333333

Watch out for integer overflows.


local n,x=io.read("n","n")
local a={}
for i=1,n do
    a[i]=io.read("n")
end

local counter=0
for i=1,n-1 do
    local sum=a[i]+a[i+1]
    if sum>x then
        local diff=sum-x
        counter=counter+diff
        if diff<a[i+1] then
            a[i+1]=a[i+1]-diff
        else
            diff=diff-a[i+1]
            a[i+1]=0
            a[i]=a[i]-diff
        end
    end
end
print(string.format("%d",counter))

Score : 300 points

Problem Statement

There are N boxes arranged in a row.
Initially, the i-th box from the left contains a_i candies.

Snuke can perform the following operation any number of times:

Choose a box containing at least one candy, and eat one of the candies in the chosen box.

His objective is as follows:

Any two neighboring boxes contain at most x candies in total.

Find the minimum number of operations required to achieve the objective.

Constraints

2 ≤ N ≤ 10^5

0 ≤ a_i ≤ 10^9

0 ≤ x ≤ 10^9

Input

The input is given from Standard Input in the following format:

N x
a_1 a_2 ... a_N

Output

Print the minimum number of operations required to achieve the objective.

Sample Input 1

3 3
2 2 2

Sample Output 1

1

Eat one candy in the second box.
Then, the number of candies in each box becomes (2, 1, 2).

Sample Input 2

6 1
1 6 1 2 0 4

Sample Output 2

11

For example, eat six candies in the second box, two in the fourth box, and three in the sixth box.
Then, the number of candies in each box becomes (1, 0, 1, 0, 0, 1).

Sample Input 3

5 9
3 1 4 1 5

Sample Output 3

0

The objective is already achieved without performing operations.

Sample Input 4

2 0
5 5

Sample Output 4

10

All the candies need to be eaten.


s = io.read()
f = s:sub(1, 1) == s:sub(#s, #s)
g = #s % 2 == 0
print(f == g and "First" or "Second")

Score : 500 points

Problem Statement

There is a string s of length 3 or greater.
No two neighboring characters in s are equal.

Takahashi and Aoki will play a game against each other.
The two players alternately performs the following operation, Takahashi going first:

Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.

The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Constraints

3 ≤ |s| ≤ 10^5

s consists of lowercase English letters.

No two neighboring characters in s are equal.

Input

The input is given from Standard Input in the following format:

s

Output

If Takahashi will win, print First. If Aoki will win, print Second.

Sample Input 1

aba

Sample Output 1

Second

Takahashi, who goes first, cannot perform the operation, since removal of the b, which is the only character not at either ends of s, would result in s becoming aa, with two as neighboring.

Sample Input 2

abc

Sample Output 2

First

When Takahashi removes b from s, it becomes ac.
Then, Aoki cannot perform the operation, since there is no character in s, excluding both ends.

Sample Input 3

abcab

Sample Output 3

First


local mce, mfl, msq, mmi, mma = math.ceil, math.floor, math.sqrt, math.min, math.max
local xs, ys, xt, yt = io.read("*n", "*n", "*n", "*n")
local n = io.read("*n")
local x, y, r = {}, {}, {}
x[1], y[1], r[1] = xs, ys, 0
for i = 1, n do
  x[i + 1], y[i + 1], r[i + 1] = io.read("*n", "*n", "*n")
end
x[n + 2], y[n + 2], r[n + 2] = xt, yt, 0
local n2 = n + 2
local len = {}
len[1] = 0
for i = 2, n2 do
  len[i] = 10000000000
end
local costs = {}
for i = 1, n2 * n2 do
  costs[i] = 10000000000
end

local function getcost(idx1, idx2)
  return mma(0, msq((x[idx1] - x[idx2]) * (x[idx1] - x[idx2]) + (y[idx1] - y[idx2]) * (y[idx1] - y[idx2])) - r[idx1] - r[idx2])
end

for i = 1, n + 1 do
  for j = i + 1, n + 2 do
    local c = getcost(i, j)
    costs[(i - 1) * n2 + j] = c
    costs[(j - 1) * n2 + i] = c
  end
end

local taskstate = {}
for i = 1, n2 do taskstate[i] = false end
local tasks = {}
local tasknum = 0
local done = 0
local tasklim = n2

local function addtask(idx)
  taskstate[idx] = true
  tasknum = tasknum + 1
  local taskidx = tasknum % tasklim
  if taskidx == 0 then taskidx = tasklim end
  tasks[taskidx] = idx
end
addtask(1)

while done < tasknum do
  done = done + 1
  local taskidx = done % tasklim
  if taskidx == 0 then taskidx = tasklim end
  local src = tasks[taskidx]
  taskstate[src] = false
  local at = {}
  for dst = 1, n2 do
    local c = costs[(src - 1) * n2 + dst]
    if len[src] + c < len[dst] then
      len[dst] = len[src] + c
      if dst ~= n2 then
        if not taskstate[dst] then
          table.insert(at, dst)
        end
      end
    end
  end
  table.sort(at, function(x, y) return len[x] < len[y] end)
  for i = 1, #at do addtask(at[i]) end
end
print(string.format("%.9f", len[n2]))

Score : 600 points

Problem Statement

On the xy-plane, Snuke is going to travel from the point (x_s, y_s) to the point (x_t, y_t).
He can move in arbitrary directions with speed 1.
Here, we will consider him as a point without size.

There are N circular barriers deployed on the plane.
The center and the radius of the i-th barrier are (x_i, y_i) and r_i, respectively.
The barriers may overlap or contain each other.

A point on the plane is exposed to cosmic rays if the point is not within any of the barriers.

Snuke wants to avoid exposure to cosmic rays as much as possible during the travel.
Find the minimum possible duration of time he is exposed to cosmic rays during the travel.

Constraints

All input values are integers.

-10^9 ≤ x_s, y_s, x_t, y_t ≤ 10^9

(x_s, y_s) ≠ (x_t, y_t)

1≤N≤1,000

-10^9 ≤ x_i, y_i ≤ 10^9

1 ≤ r_i ≤ 10^9

Input

The input is given from Standard Input in the following format:

x_s y_s x_t y_t
N
x_1 y_1 r_1
x_2 y_2 r_2
:
x_N y_N r_N

Output

Print the minimum possible duration of time Snuke is exposed to cosmic rays during the travel.
The output is considered correct if the absolute or relative error is at most 10^{-9}.

Sample Input 1

-2 -2 2 2
1
0 0 1

Sample Output 1

3.6568542495

An optimal route is as follows:

Sample Input 2

-2 0 2 0
2
-1 0 2
1 0 2

Sample Output 2

0.0000000000

An optimal route is as follows:

Sample Input 3

4 -2 -2 4
3
0 0 2
4 0 1
0 4 1

Sample Output 3

4.0000000000

An optimal route is as follows:


local mod = 1000000007
local mfl = math.floor
local function bmul(x, y)
  local x0, y0 = x % 31623, y % 31623
  local x1, y1 = mfl(x / 31623), mfl(y / 31623)
  return (x1 * y1 * 14122 + (x1 * y0 + x0 * y1) * 31623 + x0 * y0) % mod
end
local n = io.read("*n")
local t = {}
for i = 1, n do t[i] = {io.read("*n"), 1} end
for i = 1, n do t[i + n] = {io.read("*n"), 2} end
table.sort(t, function(x, y) return x[1] < y[1] end)
local ret = 1
local state = 0
for i = 1, 2 * n do
  local cur = t[i]
  if cur[2] == 1 then
    if 0 <= state then
      state = state + 1
    else
      ret = bmul(ret, -state)
      state = state + 1
    end
  else
    if state <= 0 then
      state = state - 1
    else
      ret = bmul(ret, state)
      state = state - 1
    end
  end
end
print(ret)

Score : 500 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

There are N computers and N sockets in a one-dimensional world.
The coordinate of the i-th computer is a_i, and the coordinate of the i-th socket is b_i.
It is guaranteed that these 2N coordinates are pairwise distinct.

Snuke wants to connect each computer to a socket using a cable.
Each socket can be connected to only one computer.

In how many ways can he minimize the total length of the cables?
Compute the answer modulo 10^9+7.

Constraints

1 ≤ N ≤ 10^5

0 ≤ a_i, b_i ≤ 10^9

The coordinates are integers.

The coordinates are pairwise distinct.

Input

The input is given from Standard Input in the following format:

N
a_1
:
a_N
b_1
:
b_N

Output

Print the number of ways to minimize the total length of the cables, modulo 10^9+7.

Sample Input 1

2
0
10
20
30

Sample Output 1

2

There are two optimal connections: 0-20, 10-30 and 0-30, 10-20.
In both connections the total length of the cables is 40.

Sample Input 2

3
3
10
8
7
12
5

Sample Output 2

1


local mce, mfl, msq, mmi, mma = math.ceil, math.floor, math.sqrt, math.min, math.max

local function getxor(x, y)
  local ret = 0
  local mul = 1
  while(0 < x or 0 < y) do
    if((x % 2) + (y % 2) == 1) then
      ret = ret + mul
    end
    x, y, mul = mfl(x / 2), mfl(y / 2), mul * 2
  end
  return ret
end

local n = io.read("*n")
local tot = 0
local cutpos = {}
for i = 1, 32 do cutpos[i] = false end
for i = 1, n do
  local a = io.read("*n")
  tot = getxor(tot, a)
  local pos = 1
  while a % 2 == 0 do
    pos = pos + 1
    a = mfl(a / 2)
  end
  cutpos[pos] = true
end
local tgt = {}
local isok = true
while 0 < tot do
  table.insert(tgt, tot % 2 == 1)
  tot = mfl(tot / 2)
end
local cnt = 0
for i = #tgt, 1, -1 do
  if tgt[i] then
    if cutpos[i] then
      cnt = cnt + 1
      for j = 1, i - 1 do
        tgt[j] = not tgt[j]
      end
    else
      isok = false
      break
    end
  end
end
print(isok and cnt or -1)

Score : 500 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

A cheetah and a cheater are going to play the game of Nim.
In this game they use N piles of stones.
Initially the i-th pile contains a_i stones.
The players take turns alternately, and the cheetah plays first.
In each turn, the player chooses one of the piles, and takes one or more stones from the pile.
The player who can't make a move loses.

However, before the game starts, the cheater wants to cheat a bit to make sure that he can win regardless of the moves by the cheetah.
From each pile, the cheater takes zero or one stone and eats it before the game.
In case there are multiple ways to guarantee his winning, he wants to minimize the number of stones he eats.

Compute the number of stones the cheater will eat.
In case there is no way for the cheater to win the game even with the cheating, print -1 instead.

Constraints

1 ≤ N ≤ 10^5

2 ≤ a_i ≤ 10^9

Input

The input is given from Standard Input in the following format:

N
a_1
:
a_N

Output

Print the answer.

Sample Input 1

3
2
3
4

Sample Output 1

3

The only way for the cheater to win the game is to take stones from all piles and eat them.

Sample Input 2

3
100
100
100

Sample Output 2

-1


local function getgcd(x, y)
  while 0 < x do
    x, y = y % x, x
  end
  return y
end
local a, b = io.read("*n", "*n")
local c, d = io.read("*n", "*n")
a = math.abs(a - c)
b = math.abs(b - d)
local gcd = getgcd(a, b)
a, b = a // gcd, b // gcd
print(gcd * (a + b - 1))

Score : 100 points

Problem Statement

Takahashi is drawing a segment on grid paper.

From a certain square, a square that is x squares to the right and y squares above, is denoted as square (x, y).

When Takahashi draws a segment connecting the lower left corner of square (A, B) and the lower left corner of square (C, D), find the number of the squares crossed by the segment.

Here, the segment is said to cross a square if the segment has non-empty intersection with the region within the square, excluding the boundary.

Constraints

1 \leq A, B, C, D \leq 10^9

At least one of A \neq C and B \neq D holds.

Input

The input is given from Standard Input in the following format:

A B C D

Output

Print the number of the squares crossed by the segment.

Sample Input 1

1 1 3 4

Sample Output 1

4

Sample Input 2

2 3 10 7

Sample Output 2

8


local bls, brs = bit.lshift, bit.rshift
local n = io.read("*n")
local t = {}

local lastval = 10000
while 0 < n do
  local min, max = 0, lastval
  while 1 < max - min do
    local mid = brs(min + max, 1)
    if n <= brs(mid * (mid + 1), 1) then
      max = mid
    else
      min = mid
    end
  end
  lastval = max
  table.insert(t, max)
  n = n - max
end
print(table.concat(t, "\n"))

Score : 300 points

Problem Statement

The problem set at CODE FESTIVAL 20XX Finals consists of N problems.

The score allocated to the i-th (1≦i≦N) problem is i points.

Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve.

As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him.

Determine the set of problems that should be solved.

Constraints

1≦N≦10^7

Partial Score

200 points will be awarded for passing the test set satisfying 1≦N≦1000.

Additional 100 points will be awarded for passing the test set without additional constraints.

Input

The input is given from Standard Input in the following format:

N

Output

Among the sets of problems with the total score of N, find a set in which the highest score of a problem is minimum, then print the indices of the problems in the set in any order, one per line.

If there exists more than one such set, any of them will be accepted.

Sample Input 1

4

Sample Output 1

1
3

Solving only the 4-th problem will also result in the total score of 4 points, but solving the 1-st and 3-rd problems will lower the highest score of a solved problem.

Sample Input 2

7

Sample Output 2

1
2
4

The set \{3,4\} will also be accepted.

Sample Input 3

1

Sample Output 3

1


local n=io.read("n")

local function upper_bound(x)
    local max=n+1
    local min=0
    while 1<max-min do
        mid=(max+min)//2
        if mid*(mid+1)//2>=n then
            max=mid
        else
            min=mid
        end
    end
    return max
end

local x=upper_bound(n)
for i=1,x do
    if i~=x*(x+1)//2-n then
        print(i)
    end
end


local function uf_initialize(n)
  local parent = {}
  for i = 1, n do parent[i] = i end
  return parent
end

local function uf_findroot(idx, parent)
  local idx_update = idx
  while parent[idx] ~= idx do
    idx = parent[idx]
  end
  while parent[idx_update] ~= idx do
    parent[idx_update], idx_update = idx, parent[idx_update]
  end
  return idx
end

local n, m = io.read("*n", "*n")
local lang = {}
for i = 1, m do
  lang[i] = {}
end
for i = 1, n do
  local k = io.read("*n")
  for j = 1, k do
    local l = io.read("*n")
    table.insert(lang[l], i)
  end
end

local parent = uf_initialize(n)
for i = 1, m do
  if 0 < #lang[i] then
    local r = uf_findroot(lang[i][1], parent)
    for j = 2, #lang[i] do
      local tgt = lang[i][j]
      local tr = uf_findroot(tgt, parent)
      parent[tr], parent[tgt] = r, r
    end
  end
end
local cand = uf_findroot(1, parent)
for i = 2, n do
  local z = uf_findroot(i, parent)
  if z ~= cand then print("NO") os.exit() end
end
print("YES")

Score : 400 points

Problem Statement

On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M.

For CODE FESTIVAL 20XX held on this planet, N participants gathered from all over the planet.

The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}.

Two participants A and B can communicate with each other if and only if one of the following conditions is satisfied:

There exists a language that both A and B can speak.

There exists a participant X that both A and B can communicate with.

Determine whether all N participants can communicate with all other participants.

Constraints

2≦N≦10^5

1≦M≦10^5

1≦K_i≦M

(The sum of all K_i)≦10^5

1≦L_{i,j}≦M

L_{i,1}, L_{i,2}, ..., L_{i,{}K_i} are pairwise distinct.

Partial Score

200 points will be awarded for passing the test set satisfying the following: N≦1000, M≦1000 and (The sum of all K_i)≦1000.

Additional 200 points will be awarded for passing the test set without additional constraints.

Input

The input is given from Standard Input in the following format:

N M
K_1 L_{1,1} L_{1,2} ... L_{1,{}K_1}
K_2 L_{2,1} L_{2,2} ... L_{2,{}K_2}
:
K_N L_{N,1} L_{N,2} ... L_{N,{}K_N}

Output

If all N participants can communicate with all other participants, print YES. Otherwise, print NO.

Sample Input 1

4 6
3 1 2 3
2 4 2
2 4 6
1 6

Sample Output 1

YES

Any two participants can communicate with each other, as follows:

Participants 1 and 2: both can speak language 2.

Participants 2 and 3: both can speak language 4.

Participants 1 and 3: both can communicate with participant 2.

Participants 3 and 4: both can speak language 6.

Participants 2 and 4: both can communicate with participant 3.

Participants 1 and 4: both can communicate with participant 2.

Note that there can be languages spoken by no participant.

Sample Input 2

4 4
2 1 2
2 1 2
1 3
2 4 3

Sample Output 2

NO

For example, participants 1 and 3 cannot communicate with each other.


local n, m = io.read("*n", "*n")
local t = {}
local mod = {}
local pr = {}
for i = 1, m do mod[i] = 0  pr[i] = 0 end
for i = 1, n do
  local a = io.read("*n")
  local v = a % m
  if v == 0 then v = m end
  mod[v] = mod[v] + 1
  t[i] = a
end
table.sort(t)
local pos = 1
while pos <= n - 1 do
  if t[pos] == t[pos + 1] then
    local v = t[pos] % m
    if v == 0 then v = m end
    pr[v] = pr[v] + 1
    pos = pos + 2
  else
    pos = pos + 1
  end
end
local mmi, mfl = math.min, math.floor
local ret = 0
local hm = mfl((m - 1) / 2)
for i = 1, hm do
  local v = mmi(mod[i], mod[m - i])
  ret = ret + v
  mod[i] = mod[i] - v
  mod[m - i] = mod[m - i] - v
end
do
  local v = mfl(mod[m] / 2)
  ret = ret + v
  mod[m] = mod[m] - v * 2
end
if m % 2 == 0 then
  local z = mfl(m / 2)
  local v = mfl(mod[z] / 2)
  ret = ret + v
  mod[z] = mod[z] - v * 2
end
for i = 1, m do
  local v = mmi(pr[i], mfl(mod[i] / 2))
  ret = ret + v
end
print(ret)

Score : 700 points

Problem Statement

Takahashi is playing with N cards.

The i-th card has an integer X_i on it.

Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:

The integers on the two cards are the same.

The sum of the integers on the two cards is a multiple of M.

Find the maximum number of pairs that can be created.

Note that a card cannot be used in more than one pair.

Constraints

2≦N≦10^5

1≦M≦10^5

1≦X_i≦10^5

Input

The input is given from Standard Input in the following format:

N M
X_1 X_2 ... X_N

Output

Print the maximum number of pairs that can be created.

Sample Input 1

7 5
3 1 4 1 5 9 2

Sample Output 1

3

Three pairs (3,2), (1,4) and (1,9) can be created.

It is possible to create pairs (3,2) and (1,1), but the number of pairs is not maximized with this.

Sample Input 2

15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99

Sample Output 2

6


local mmi, mma = math.min, math.max
local MinCostFlow = {}

MinCostFlow.initialize = function(self, n, spos, tpos, inf)
  self.n = n
  self.spos, self.tpos = spos, tpos
  self.inf = inf
  -- edge_dst[src][i] := dst
  self.edge_dst = {}
  -- edge_cost[src][i] := cost from src to edge_dst[src][i]
  self.edge_cost = {}
  -- edge_cap[src][i] := capacity from src to edge_dst[src][i]
  self.edge_cap = {}
  -- initial capacity. corresponding to edge_cap
  self.edge_initialcap = {}
  -- edge_dst_invedge_idx[src][i] := "j" where edge_dst[dst][j] == src
  -- in this case, edge_dst_invedge_idx[dst][j] should be "i".
  self.edge_dst_invedge_idx = {}
  -- len[v] := length from spos. len[spos] := 0
  self.len = {}
  -- sub_graph_flag[v] := whether to contains the vertex v in the sub-graph or not
  self.sub_graph_flag = {}
  -- sub_graph_v[i] := list of vertexes that are contained in the sub-graph. from tpos to spos.
  self.sub_graph_v = {}
  -- sub_graph_edgeidx[i] := edge index from sub_graph_v[i + 1] to sub_graph_v[i]
  self.sub_graph_edgeidx = {}
  -- sub_graph_size := the size of sub_graph_v.
  -- may not equal to #sub_graph_v (because not cleared).
  self.sub_graph_size = 0
  for i = 1, n do
    self.edge_dst[i] = {}
    self.edge_cost[i] = {}
    self.edge_cap[i] = {}
    self.edge_initialcap[i] = {}
    self.edge_dst_invedge_idx[i] = {}
    self.len[i] = 0
    self.sub_graph_flag[i] = false
  end
end

MinCostFlow.addEdge = function(self, src, dst, cost, cap)
  table.insert(self.edge_dst[src], dst)
  table.insert(self.edge_cost[src], cost)
  table.insert(self.edge_cap[src], cap)
  table.insert(self.edge_initialcap[src], cap)
  table.insert(self.edge_dst_invedge_idx[src], 1 + #self.edge_dst[dst])
  table.insert(self.edge_dst[dst], src)
  table.insert(self.edge_cost[dst], -cost)
  table.insert(self.edge_cap[dst], 0)--invcap
  table.insert(self.edge_initialcap[dst], 0)--invcap
  table.insert(self.edge_dst_invedge_idx[dst], #self.edge_dst[src])
end
MinCostFlow.invwalk_recursive = function(self, invsrc)
  if invsrc == self.spos then return true end
  local edge_dst, edge_cap, edge_cost = self.edge_dst, self.edge_cap, self.edge_cost
  local edge_dst_invedge_idx = self.edge_dst_invedge_idx
  local len = self.len
  local sub_graph_flag = self.sub_graph_flag
  local sub_graph_v = self.sub_graph_v
  local sub_graph_edgeidx = self.sub_graph_edgeidx
  for i = 1, #edge_dst[invsrc] do
    local invdst = edge_dst[invsrc][i]
    local j = edge_dst_invedge_idx[invsrc][i]
    if 0 < edge_cap[invdst][j]
    and len[invdst] + edge_cost[invdst][j] == len[invsrc]
    and not sub_graph_flag[invdst] then
      self.sub_graph_size = self.sub_graph_size + 1
      sub_graph_v[self.sub_graph_size] = invdst
      sub_graph_edgeidx[self.sub_graph_size - 1] = j
      sub_graph_flag[invdst] = true
      if self:invwalk_recursive(invdst) then
        return true
      else
        self.sub_graph_flag[invdst] = false
        self.sub_graph_size = self.sub_graph_size - 1
      end
    end
  end
  return false
end

MinCostFlow.makeSubGraph = function(self)
  local inf = self.inf
  local len = self.len
  local n = self.n
  local edge_dst, edge_cap = self.edge_dst, self.edge_cap
  local edge_dst_invedge_idx = self.edge_dst_invedge_idx
  local edge_cost = self.edge_cost
  local sub_graph_v = self.sub_graph_v
  local sub_graph_edgeidx = self.sub_graph_edgeidx
  local sub_graph_flag = self.sub_graph_flag
  for i = 1, self.n do
    len[i] = inf
    sub_graph_flag[i] = false
  end
  -- Bellman-Ford
  local updated1, updated2 = {}, {}
  for i = 1, n do
    updated1[i] = true
  end
  len[self.spos] = 0
  for irp = 1, n do
    local updsrc = irp % 2 == 1 and updated1 or updated2
    local upddst = irp % 2 == 1 and updated2 or updated1
    for i = 1, n do
      upddst[i] = false
    end
    for src = 1, n do
      if updsrc[src] then
        local eddst, edcap, edcost = edge_dst[src], edge_cap[src], edge_cost[src]
        for i = 1, #eddst do
          if 0 < edcap[i] then
            local dst = eddst[i]
            local cost = edcost[i]
            if len[src] + cost < len[dst] then
              len[dst] = len[src] + cost
              upddst[dst] = true
            end
          end
        end
      end
    end
  end
  self.sub_graph_size = 0
  if inf <= len[self.tpos] then
    return 0
  end
  -- restore route (from tpos to spos)
  self.sub_graph_size = 1
  sub_graph_v[1] = self.tpos
  self:invwalk_recursive(self.tpos)
  local min_capacity = inf
  for i = self.sub_graph_size, 2, -1 do
    local src = sub_graph_v[i]
    local j = sub_graph_edgeidx[i - 1]
    min_capacity = mmi(min_capacity, edge_cap[src][j])
  end
  return min_capacity
end

MinCostFlow.flow = function(self, capacity)
  local edge_dst, edge_cap = self.edge_dst, self.edge_cap
  local edge_dst_invedge_idx = self.edge_dst_invedge_idx
  local sub_graph_v = self.sub_graph_v
  local sub_graph_edgeidx = self.sub_graph_edgeidx
  for i = self.sub_graph_size, 2, -1 do
    local src = sub_graph_v[i]
    local dst = sub_graph_v[i - 1]
    local j = sub_graph_edgeidx[i - 1]
    edge_cap[src][j] = edge_cap[src][j] - capacity
    local k = edge_dst_invedge_idx[src][j]
    edge_cap[dst][k] = edge_cap[dst][k] + capacity
  end
end

MinCostFlow.getMinCostFlow = function(self, amount, invalid)
  local ret = 0
  local cap = self:makeSubGraph()
  while 0 < cap do
    cap = mmi(amount, cap)
    ret = ret + self.len[self.tpos] * cap
    self:flow(cap)
    amount = amount - cap
    if 0 < amount then
      cap = self:makeSubGraph()
    else
      break
    end
  end
  if 0 < amount then return invalid end
  return ret
end

local h, w = io.read("*n", "*n")
local s = 2 * h * w + 1
local t = 2 * h * w + 2
MinCostFlow:initialize(t, s, t, 1000000007 * 100)
MinCostFlow:addEdge(s, 1, 0, 2)
MinCostFlow:addEdge(2 * h * w, t, 0, 2)
for i = 1, h do
  for j = 1, w - 1 do
    local idx = (i - 1) * w + j
    MinCostFlow:addEdge(idx + h * w, idx + 1, 0, 2)
  end
end
for i = 1, h - 1 do
  for j = 1, w do
    local idx = (i - 1) * w + j
    MinCostFlow:addEdge(idx + h * w, idx + w, 0, 2)
  end
end
for i = 1, h do
  for j = 1, w do
    local idx = (i - 1) * w + j
    local a = io.read("*n")
    MinCostFlow:addEdge(idx, idx + h * w, 0, 1)
    MinCostFlow:addEdge(idx, idx + h * w, -a, 1)
  end
end
print(math.abs(MinCostFlow:getMinCostFlow(2)))

Max Score: $600$ Points

Problem Statement

Sigma and his brother Sugim are in the $H \times W$ grid. They wants to buy some souvenirs.

Their start position is upper-left cell, and the goal position is lower-right cell.

Some cells has a souvenir shop. At $i$-th row and $j$-th column, there is $a_{i, j}$ souvenirs.

In one move, they can go left, right, down, and up cell.

But they have little time, so they can move only $H+W-2$ times.

They wanted to buy souvenirs as many as possible, but they had no computer, so they couldn't get the maximal numbers of souvenirs.

Write a program and calculate the maximum souvenirs they can get, and help them.

Input

The input is given from standard input in the following format.

$H \ W$
$a_{1, 1} \ a_{1, 2} \ \cdots \ a_{1, W}$
$a_{2, 1} \ a_{2, 2} \ \cdots \ a_{2, W}$
$\vdots \ \ \ \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \vdots$
$a_{H, 1} \ a_{H, 2} \ \cdots \ a_{H, W}$

Output

Print the maximum number of souvenirs they can get.

Constraints

$1 \le H, W \le 200$

$0 \le a_{i, j} \le 10^5$

Subtasks

Subtask 1 [ 50 points ]

The testcase in the subtask satisfies $1 \le H \le 2$.

Subtask 2 [ 80 points ]

The testcase in the subtask satisfies $1 \le H \le 3$.

Subtask 3 [ 120 points ]

The testcase in the subtask satisfies $1 \le H, W \le 7$.

Subtask 4 [ 150 points ]

The testcase in the subtask satisfies $1 \le H, W \le 30$.

Subtask 5 [ 200 points ]

There are no additional constraints.

Sample Input 1

3 3
1 0 5
2 2 3
4 2 4

Sample Output 1

21

The cell at $i$-th row and $j$-th column is denoted $(i, j)$.

In this case, one of the optimal solution is this:

Sigma moves $(1, 1) -> (1, 2) -> (1, 3) -> (2, 3) -> (3, 3)$.

Sugim moves $(1, 1) -> (2, 1) -> (3, 1) -> (3, 2) -> (3, 3)$.

Then, they can get $21$ souvernirs.

Sample Input 2

6 6
1 2 3 4 5 6
8 6 9 1 2 0
3 1 4 1 5 9
2 6 5 3 5 8
1 4 1 4 2 1
2 7 1 8 2 8

Sample Output 2

97

Writer : square1001


local h,w=io.read("*n","*n","*l")
local counter=0
for i=1,h do
    local a=#io.read():gsub("%.","")
    counter=counter+a
end
print(counter==h+w-1 and "Possible" or "Impossible")

Score : 200 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

We have a grid of H rows and W columns. Initially, there is a stone in the top left cell. Shik is trying to move the stone to the bottom right cell. In each step, he can move the stone one cell to its left, up, right, or down (if such cell exists). It is possible that the stone visits a cell multiple times (including the bottom right and the top left cell).

You are given a matrix of characters a_{ij} (1 \leq i \leq H, 1 \leq j \leq W). After Shik completes all moving actions, a_{ij} is # if the stone had ever located at the i-th row and the j-th column during the process of moving. Otherwise, a_{ij} is .. Please determine whether it is possible that Shik only uses right and down moves in all steps.

Constraints

2 \leq H, W \leq 8

a_{i,j} is either # or ..

There exists a valid sequence of moves for Shik to generate the map a.

Input

The input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If it is possible that Shik only uses right and down moves, print Possible. Otherwise, print Impossible.

Sample Input 1

4 5
##...
.##..
..##.
...##

Sample Output 1

Possible

The matrix can be generated by a 7-move sequence: right, down, right, down, right, down, and right.

Sample Input 2

5 3
###
..#
###
#..
###

Sample Output 2

Impossible

Sample Input 3

4 5
##...
.###.
.###.
...##

Sample Output 3

Impossible


local n = io.read("*n")
local a, b = {}, {}
for i = 1, n do
  a[i] = 30000 * i
  b[i] = 30000 * (n + 1 - i)
end
for i = 1, n do
  local p = io.read("*n")
  a[p] = a[p] + i
end
print(table.concat(a, " "))
print(table.concat(b, " "))

Score : 400 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

You are given a permutation p of the set {1, 2, ..., N}. Please construct two sequences of positive integers a_1, a_2, ..., a_N and b_1, b_2, ..., b_N satisfying the following conditions:

1 \leq a_i, b_i \leq 10^9 for all i

a_1 < a_2 < ... < a_N

b_1 > b_2 > ... > b_N

a_{p_1}+b_{p_1} < a_{p_2}+b_{p_2} < ... < a_{p_N}+b_{p_N}

Constraints

2 \leq N \leq 20,000

p is a permutation of the set {1, 2, ..., N}

Input

The input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

The output consists of two lines. The first line contains a_1, a_2, ..., a_N seperated by a space. The second line contains b_1, b_2, ..., b_N seperated by a space.

It can be shown that there always exists a solution for any input satisfying the constraints.

Sample Input 1

2
1 2

Sample Output 1

1 4
5 4

a_1 + b_1 = 6 and a_2 + b_2 = 8. So this output satisfies all conditions.

Sample Input 2

3
3 2 1

Sample Output 2

1 2 3
5 3 1

Sample Input 3

3
2 3 1

Sample Output 3

5 10 100
100 10 1


Score : 100 points

Problem Statement

Two students of AtCoder Kindergarten are fighting over candy packs.

There are three candy packs, each of which contains a, b, and c candies, respectively.

Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.

Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Constraints

1 ≦ a, b, c ≦ 100

Input

The input is given from Standard Input in the following format:

a b c

Output

If it is possible to distribute the packs so that each student gets the same number of candies, print Yes. Otherwise, print No.

Sample Input 1

10 30 20

Sample Output 1

Yes

Give the pack with 30 candies to one student, and give the two packs with 10 and 20 candies to the other. Then, each gets 30 candies.

Sample Input 2

30 30 100

Sample Output 2

No

In this case, the student who gets the pack with 100 candies always has more candies than the other.

Note that every pack must be given to one of them.

Sample Input 3

56 25 31

Sample Output 3

Yes


local max = math.max(arg[1], arg[2], arg[3])
local sum = arg[1] + arg[2] + arg[3]

if sum - max == max then
	print("Yes")
else
	print("No")
end


local read = io.read

local W, H, N = read("*n", "*n", "*n")
local rectangle = {}
for x = 1, W do
	rectangle[x] = {}
	for y = 1, H do
		rectangle[x][y] = true
	end
end

local function paint(x_i, y_i, a_i)
	if a_i == 1 then
		for x = 1, x_i do
			for y = 1, H do
				rectangle[x][y] = false
			end
		end
	elseif a_i == 2 then
		for x = x_i + 1, W do
			for y = 1, H do
				rectangle[x][y] = false
			end
		end
	elseif a_i == 3 then
		for x = 1, W do
			for y = 1, y_i do
				rectangle[x][y] = false
			end
		end
	else
		for x = 1, W do
			for y = y_i + 1, H do
				rectangle[x][y] = false
			end
		end
	end
end

for i = 1, N do
	paint(read("*n", "*n", "*n"))
end

local out = 0
for x = 1, W do
	for y = 1, H do
		if rectangle[x][y] then
			out = out + 1
		end
	end
end

print(out)

Score : 200 points

Problem Statement

There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white.

Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i).

Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows:

If a_i = 1, he painted the region satisfying x < x_i within the rectangle.

If a_i = 2, he painted the region satisfying x > x_i within the rectangle.

If a_i = 3, he painted the region satisfying y < y_i within the rectangle.

If a_i = 4, he painted the region satisfying y > y_i within the rectangle.

Find the area of the white region within the rectangle after he finished painting.

Constraints

1 ≦ W, H ≦ 100

1 ≦ N ≦ 100

0 ≦ x_i ≦ W (1 ≦ i ≦ N)

0 ≦ y_i ≦ H (1 ≦ i ≦ N)

W, H (21:32, added), x_i and y_i are integers.

a_i (1 ≦ i ≦ N) is 1, 2, 3 or 4.

Input

The input is given from Standard Input in the following format:

W H N
x_1 y_1 a_1
x_2 y_2 a_2
:
x_N y_N a_N

Output

Print the area of the white region within the rectangle after Snuke finished painting.

Sample Input 1

5 4 2
2 1 1
3 3 4

Sample Output 1

9

The figure below shows the rectangle before Snuke starts painting.

First, as (x_1, y_1) = (2, 1) and a_1 = 1, he paints the region satisfying x < 2 within the rectangle:

Then, as (x_2, y_2) = (3, 3) and a_2 = 4, he paints the region satisfying y > 3 within the rectangle:

Now, the area of the white region within the rectangle is 9.

Sample Input 2

5 4 3
2 1 1
3 3 4
1 4 2

Sample Output 2

0

It is possible that the whole region within the rectangle is painted black.

Sample Input 3

10 10 5
1 6 1
4 1 3
6 9 4
9 4 2
3 1 3

Sample Output 3

64


local n, t = io.read("*n", "*n")
local a = {}
for i = 1, n do a[i] = io.read("*n") end
local mmi, mma = math.min, math.max
local min = {}
min[1] = {}
min[1].c, min[1].v = 1, a[1]

for i = 2, n do
  min[i] = {}
  if a[i] < min[i - 1].v then
    min[i].c, min[i].v = 1, a[i]
  elseif min[i - 1].v == a[i] then
    min[i].c = min[i - 1].c + 1
    min[i].v = min[i - 1].v
  else
    min[i].c = min[i - 1].c
    min[i].v = min[i - 1].v
  end
end
local maxrate = 0
for i = 2, n do
  maxrate = mma(maxrate, a[i] - min[i].v)
end
local totcnt = 0
local mincnt = min[1].c
local maxcnt = 0
for i = 2, n do
  if min[i].v < min[i - 1].v then
    totcnt = totcnt + mmi(mincnt, maxcnt)
    mincnt = min[i].c
    maxcnt = 0
  else
    if maxrate == a[i] - min[i].v then
      maxcnt = maxcnt + 1
    end
  end
end
totcnt = totcnt + mmi(mincnt, maxcnt)
print(totcnt)

Score : 400 points

Problem Statement

There are N towns located in a line, conveniently numbered 1 through N. Takahashi the merchant is going on a travel from town 1 to town N, buying and selling apples.

Takahashi will begin the travel at town 1, with no apple in his possession. The actions that can be performed during the travel are as follows:

Move: When at town i (i < N), move to town i + 1.

Merchandise: Buy or sell an arbitrary number of apples at the current town. Here, it is assumed that one apple can always be bought and sold for A_i yen (the currency of Japan) at town i (1 ≦ i ≦ N), where A_i are distinct integers. Also, you can assume that he has an infinite supply of money.

For some reason, there is a constraint on merchandising apple during the travel: the sum of the number of apples bought and the number of apples sold during the whole travel, must be at most T. (Note that a single apple can be counted in both.)

During the travel, Takahashi will perform actions so that the profit of the travel is maximized. Here, the profit of the travel is the amount of money that is gained by selling apples, minus the amount of money that is spent on buying apples. Note that we are not interested in apples in his possession at the end of the travel.

Aoki, a business rival of Takahashi, wants to trouble Takahashi by manipulating the market price of apples. Prior to the beginning of Takahashi's travel, Aoki can change A_i into another arbitrary non-negative integer A_i' for any town i, any number of times. The cost of performing this operation is |A_i - A_i'|. After performing this operation, different towns may have equal values of A_i.

Aoki's objective is to decrease Takahashi's expected profit by at least 1 yen. Find the minimum total cost to achieve it. You may assume that Takahashi's expected profit is initially at least 1 yen.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9 (1 ≦ i ≦ N)

A_i are distinct.

2 ≦ T ≦ 10^9

In the initial state, Takahashi's expected profit is at least 1 yen.

Input

The input is given from Standard Input in the following format:

N T
A_1 A_2 ... A_N

Output

Print the minimum total cost to decrease Takahashi's expected profit by at least 1 yen.

Sample Input 1

3 2
100 50 200

Sample Output 1

1

In the initial state, Takahashi can achieve the maximum profit of 150 yen as follows:

Move from town 1 to town 2.

Buy one apple for 50 yen at town 2.

Move from town 2 to town 3.

Sell one apple for 200 yen at town 3.

If, for example, Aoki changes the price of an apple at town 2 from 50 yen to 51 yen, Takahashi will not be able to achieve the profit of 150 yen. The cost of performing this operation is 1, thus the answer is 1.

There are other ways to decrease Takahashi's expected profit, such as changing the price of an apple at town 3 from 200 yen to 199 yen.

Sample Input 2

5 8
50 30 40 10 20

Sample Output 2

2

Sample Input 3

10 100
7 10 4 5 9 3 6 8 2 1

Sample Output 3

2


local mmi, mma = math.min, math.max
local n = io.read("*n")
local edge = {}
local len = {}
local parent = {}
local childcnt = {}
local valmin, valmax = {}, {}
for i = 1, n do
  edge[i] = {}
  len[i] = -1
  parent[i] = 0
  childcnt[i] = 0
  valmin[i], valmax[i] = false, false
end
for i = 1, n - 1 do
  local a, b = io.read("*n", "*n")
  edge[a][b], edge[b][a] = true, true
end
local k = io.read("*n")
local startpos, startval = 0, 0
for i = 1, k do
  local pos, val = io.read("*n", "*n")
  valmin[pos], valmax[pos] = val, val
  if i == 1 or val < startval then
    startpos, startval = pos, val
  end
end
local valid = true
len[startpos] = 0
local tasks = {startpos}
for i = 1, n do
  local src = tasks[i]
  for dst, _u in pairs(edge[src]) do
    if len[dst] < 0 then
      childcnt[src] = childcnt[src] + 1
      parent[dst] = src
      len[dst] = len[src] + 1
      if valmax[dst] then
        if (valmax[dst] + valmax[startpos] + len[dst]) % 2 == 1 then
          valid = false break
        end
        if valmax[startpos] + len[dst] < valmax[dst] then
          valid = false break
        end
      else
        valmax[dst] = valmax[startpos] + len[dst]
        valmin[dst] = valmax[startpos] + (len[dst] % 2)
      end
      table.insert(tasks, dst)
    end
  end
  if not valid then break end
end
if not valid then print("No")
else
  tasks = {}
  for i = 1, n do
    if childcnt[i] == 0 then
      table.insert(tasks, i)
    end
  end
  for i = 1, n - 1 do
    local c = tasks[i]
    local p = parent[c]
    local newp_min = mma(valmin[p], valmin[c] - 1)
    local newp_max = mmi(valmax[p], valmax[c] + 1)
    if newp_max < newp_min then
      valid = false
      break
    end
    valmin[p] = newp_min
    valmax[p] = newp_max
    childcnt[p] = childcnt[p] - 1
    if childcnt[p] == 0 then
      table.insert(tasks, p)
    end
  end
  if not valid then print("No")
  else
    local val = {}
    for i = 1, n do val[i] = 0 end
    val[startpos] = valmin[startpos]
    tasks = {startpos}
    for i = 1, n do
      local src = tasks[i]
      for dst, _u in pairs(edge[src]) do
        if len[src] < len[dst] then
          if valmin[dst] <= val[src] - 1 then
            val[dst] = val[src] - 1
          else
            val[dst] = val[src] + 1
          end
          table.insert(tasks, dst)
        end
      end
    end
    print("Yes")
    for i = 1, n do
      print(val[i])
    end
  end
end

Score : 800 points

Problem Statement

We have a tree with N vertices. The vertices are numbered 1, 2, ..., N. The i-th (1 ≦ i ≦ N - 1) edge connects the two vertices A_i and B_i.

Takahashi wrote integers into K of the vertices. Specifically, for each 1 ≦ j ≦ K, he wrote the integer P_j into vertex V_j. The remaining vertices are left empty. After that, he got tired and fell asleep.

Then, Aoki appeared. He is trying to surprise Takahashi by writing integers into all empty vertices so that the following condition is satisfied:

Condition: For any two vertices directly connected by an edge, the integers written into these vertices differ by exactly 1.

Determine if it is possible to write integers into all empty vertices so that the condition is satisfied. If the answer is positive, find one specific way to satisfy the condition.

Constraints

1 ≦ N ≦ 10^5

1 ≦ K ≦ N

1 ≦ A_i, B_i ≦ N (1 ≦ i ≦ N - 1)

1 ≦ V_j ≦ N (1 ≦ j ≦ K) (21:18, a mistake in this constraint was corrected)

0 ≦ P_j ≦ 10^5 (1 ≦ j ≦ K)

The given graph is a tree.

All v_j are distinct.

Input

The input is given from Standard Input in the following format:

N
A_1 B_1
A_2 B_2
:
A_{N-1} B_{N-1}
K
V_1 P_1
V_2 P_2
:
V_K P_K

Output

If it is possible to write integers into all empty vertices so that the condition is satisfied, print Yes. Otherwise, print No.

If it is possible to satisfy the condition, print N lines in addition. The v-th (1 ≦ v ≦ N) of these N lines should contain the integer that should be written into vertex v. If there are multiple ways to satisfy the condition, any of those is accepted.

Sample Input 1

5
1 2
3 1
4 3
3 5
2
2 6
5 7

Sample Output 1

Yes
5
6
6
5
7

The figure below shows the tree when Takahashi fell asleep. For each vertex, the integer written beside it represents the index of the vertex, and the integer written into the vertex is the integer written by Takahashi.

Aoki can, for example, satisfy the condition by writing integers into the remaining vertices as follows:

This corresponds to Sample Output 1. Note that other outputs that satisfy the condition will also be accepted, such as:

Yes
7
6
8
7
7

Sample Input 2

5
1 2
3 1
4 3
3 5
3
2 6
4 3
5 7

Sample Output 2

No

Sample Input 3

4
1 2
2 3
3 4
1
1 0

Sample Output 3

Yes
0
-1
-2
-3

The integers written by Aoki may be negative or exceed 10^6.


local mma = math.max
local mfl, mce, mmi = math.floor, math.ceil, math.min

local AvlTree = {}
AvlTree.makenode = function(self, val, parent)
  local i = self.box[#self.box]
  if not i then i = #self.v + 1
  else table.remove(self.box)
  end
  self.v[i], self.p[i] = val, parent
  self.lc[i], self.rc[i], self.l[i], self.r[i] = 0, 0, 1, 1
  return i
end
AvlTree.create = function(self, lessthan, n)
  self.lessthan = lessthan
  self.root = 1
  self.box = {}
  for i = n + 1, 2, -1 do table.insert(self.box, i) end
  -- value, leftCount, rightCount, left, right, parent
  self.v, self.lc, self.rc, self.l, self.r, self.p = {}, {}, {}, {}, {}, {}
  for i = 1, n + 1 do
    self.v[i], self.p[i] = 0, 1
    self.lc[i], self.rc[i], self.l[i], self.r[i] = 0, 0, 1, 1
  end
end

AvlTree.recalc = function(self, i)
  local kl, kr = self.l[i], self.r[i]
  if 1 < kl then self.lc[i] = 1 + mma(self.lc[kl], self.rc[kl])
  else self.lc[i] = 0
  end
  if 1 < kr then self.rc[i] = 1 + mma(self.lc[kr], self.rc[kr])
  else self.rc[i] = 0
  end
end
AvlTree.recalcAll = function(self, i)
  while 1 < i do
    self:recalc(i)
    i = self.p[i]
  end
end

AvlTree.rotR = function(self, parent)
  local granp, child = self.p[parent], self.l[parent]
  self.r[child], self.l[parent] = parent, self.r[child]
  self.p[child], self.p[parent] = granp, child
  self.p[self.l[parent]] = parent
  if 1 < granp then
    if self.l[granp] == parent then
      self.l[granp] = child
    else
      self.r[granp] = child
    end
  else
    self.root = child
  end
end

AvlTree.rotL = function(self, parent)
  local granp, child = self.p[parent], self.r[parent]
  self.l[child], self.r[parent] = parent, self.l[child]
  self.p[child], self.p[parent] = granp, child
  self.p[self.r[parent]] = parent
  if 1 < granp then
    if self.r[granp] == parent then
      self.r[granp] = child
    else
      self.l[granp] = child
    end
  else
    self.root = child
  end
end

AvlTree.rotLR = function(self, lparent, rparent)
  local sp, sl, sr = self.p, self.l, self.r
  local granp, d = sp[rparent], sr[lparent]
  sp[lparent], sr[lparent] = d, sl[d]
  sp[rparent], sl[rparent] = d, sr[d]
  sp[sl[d]], sp[sr[d]] = lparent, rparent
  sp[d], sl[d], sr[d] = granp, lparent, rparent
  if 1 < granp then
    if sr[granp] == rparent then sr[granp] = d
    else sl[granp] = d
    end
  else self.root = d
  end
end

AvlTree.rotRL = function(self, rparent, lparent)
  local sp, sl, sr = self.p, self.l, self.r
  local granp, d = sp[lparent], sl[rparent]
  sp[rparent], sl[rparent] = d, sr[d]
  sp[lparent], sr[lparent] = d, sl[d]
  sp[sr[d]], sp[sl[d]] = rparent, lparent
  sp[d], sr[d], sl[d] = granp, rparent, lparent
  if 1 < granp then
    if sl[granp] == lparent then sl[granp] = d
    else sr[granp] = d
    end
  else self.root = d
  end
end

AvlTree.empty = function(self) return self.root <= 1 end

AvlTree.push = function(self, val)
  if self.root <= 1 then self.root = self:makenode(val, 1) return end
  local pos = self.root
  while true do
    if self.lessthan(val, self.v[pos]) then
      if 1 < self.l[pos] then
        pos = self.l[pos]
      else
        self.l[pos] = self:makenode(val, pos)
        pos = self.l[pos]
        break
      end
    else
      if 1 < self.r[pos] then
        pos = self.r[pos]
      else
        self.r[pos] = self:makenode(val, pos)
        pos = self.r[pos]
        break
      end
    end
  end
  while 1 < pos do
    local child, parent = pos, self.p[pos]
    if parent <= 1 then
      break
    end
    self:recalc(parent)
    local lcp_m_rcp = self.lc[parent] - self.rc[parent]
    if lcp_m_rcp % 2 ~= 0 then -- 1 or -1
      pos = parent
    elseif lcp_m_rcp == 2 then
      if self.lc[child] - 1 == self.rc[child] then
        self:rotR(parent)
        self:recalcAll(parent)
      else
        self:rotLR(child, parent)
        self:recalc(child)
        self:recalcAll(parent)
      end
      break
    elseif lcp_m_rcp == -2 then
      if self.rc[child] - 1 == self.lc[child] then
        self:rotL(parent)
        self:recalcAll(parent)
      else
        self:rotRL(child, parent)
        self:recalc(child)
        self:recalcAll(parent)
      end
      break
    else
      break
    end
  end
end

AvlTree.rmsub = function(self, node)
  while 1 < node do
    self:recalc(node)
    if self.lc[node] == self.rc[node] then
      node = self.p[node]
    elseif self.lc[node] + 1 == self.rc[node] then
      self:recalcAll(self.p[node])
      break
    else
      if self.lc[self.r[node]] == self.rc[self.r[node]] then
        self:rotL(node)
        self:recalcAll(node)
        break
      elseif self.lc[self.r[node]] + 1 == self.rc[self.r[node]] then
        local nr = self.r[node]
        self:rotL(node)
        self:recalc(node)
        node = nr
      else
        local nr = self.r[node]
        local nrl = self.l[nr]
        self:rotRL(nr, node)
        self:recalc(nr)
        self:recalc(node)
        node = nrl
      end
    end
  end
end

AvlTree.pop = function(self)
  local node = self.root
  while 1 < self.l[node] do
    node = self.l[node]
  end
  local v = self.v[node]
  local kp = self.p[node]
  self.p[self.r[node]] = kp
  if 1 < kp then
    self.l[kp] = self.r[node]
    self:rmsub(kp)
  else
    self.root = self.r[node]
  end
  table.insert(self.box, node)
  return v
end

AvlTree.new = function(lessthan, n)
  local obj = {}
  setmetatable(obj, {__index = AvlTree})
  obj:create(lessthan, n)
  return obj
end

local n, m = io.read("*n", "*n")
local a = {}
for i = 1, m do
  a[i] = io.read("*n")
end

local function lt(p, q)
  return a[p] > a[q]
end
local avl = AvlTree.new(lt, m)
for i = 1, m do avl:push(i) end
while true do
  local top = avl:pop()
  if avl:empty() then
    print(a[top] - 1)
    break
  else
    local nxt = avl:pop()
    a[top] = a[top] - 1
    a[nxt] = a[nxt] - 1
    if 0 < a[nxt] then
      avl:push(nxt)
      avl:push(top)
    elseif 0 < a[top] then
      avl:push(top)
    end
    if avl:empty() then print(0) break end
  end
end

Score : 200 points

Problem Statement

There are K pieces of cakes.
Mr. Takahashi would like to eat one cake per day, taking K days to eat them all.

There are T types of cake, and the number of the cakes of type i (1 ≤ i ≤ T) is a_i.

Eating the same type of cake two days in a row would be no fun,
so Mr. Takahashi would like to decide the order for eating cakes that minimizes the number of days on which he has to eat the same type of cake as the day before.

Compute the minimum number of days on which the same type of cake as the previous day will be eaten.

Constraints

1 ≤ K ≤ 10000

1 ≤ T ≤ 100

1 ≤ a_i ≤ 100

a_1 + a_2 + ... + a_T = K

Input

The input is given from Standard Input in the following format:

K T
a_1 a_2 ... a_T

Output

Print the minimum number of days on which the same type of cake as the previous day will be eaten.

Sample Input 1

7 3
3 2 2

Sample Output 1

0

For example, if Mr. Takahashi eats cakes in the order of 2, 1, 2, 3, 1, 3, 1, he can avoid eating the same type of cake as the previous day.

Sample Input 2

6 3
1 4 1

Sample Output 2

1

There are 6 cakes.
For example, if Mr. Takahashi eats cakes in the order of 2, 3, 2, 2, 1, 2, he has to eat the same type of cake (i.e., type 2) as the previous day only on the fourth day.
Since this is the minimum number, the answer is 1.

Sample Input 3

100 1
100

Sample Output 3

99

Since Mr. Takahashi has only one type of cake, he has no choice but to eat the same type of cake as the previous day from the second day and after.