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 break end
  end
end


local mma, mmi = math.max, math.min
local h, w = io.read("*n", "*n", "*l")
local map = {}
for j = 1, w do map[j] = {} end
for i = 1, h do
  local str = io.read()
  for j = 1, w do
    map[j][i] = str:sub(j, j):byte()
  end
end
local function solve(col)
  local mapl, mapr = map[col], map[col + 1]
  local cost = {}
  do
    cost[1] = {}
    local lv = mapl[1]
    for right = 1, h do
      local z = lv == mapr[right] and 1 or 0
      cost[1][right] = z
    end
  end
  for left = 2, h do
    cost[left] = {}
    local lv = mapl[left]
    do
      local z = lv == mapr[1] and 1 or 0
      cost[left][1] = z
    end
    for right = 2, h do
      local z = lv == mapr[right] and 1 or 0
      z = z + cost[left - 1][right - 1]
      cost[left][right] = z
    end
  end
  for sum = h * 2 - 1, 2, -1 do
    local dst_l_min, dst_l_max = mma(1, sum - h), mmi(h, sum - 1)
    for dst_l = dst_l_min, dst_l_max do
      local dst_r = sum - dst_l
      if dst_l == h then
        cost[h][dst_r] = cost[h][dst_r] + cost[h][dst_r + 1]
      elseif dst_r == h then
        cost[dst_l][h] = cost[dst_l][h] + cost[dst_l + 1][h]
      else
        cost[dst_l][dst_r] = cost[dst_l][dst_r] + mmi(cost[dst_l + 1][dst_r], cost[dst_l][dst_r + 1])
      end
    end
  end
  local res = cost[1][1]
  for i = 1, h do
    res = mmi(res, cost[i][1])
    res = mmi(res, cost[1][i])
  end
  return res
end

local ret = 0
for i = 1, w - 1 do
  ret = ret + solve(i)
end
print(ret)

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

Score : 800 points

Problem Statement

Mr. Takahashi has in his room an art object with H rows and W columns, made up of H \times W blocks.
Each block has a color represented by a lowercase English letter (a-z).
The color of the block at the i-th row and j-th column is c_{i,j}.

Mr. Takahashi would like to dismantle the object, finding it a bit kitschy for his tastes.
The dismantling is processed by repeating the following operation:

Choose one of the W columns and push down that column one row.
The block at the bottom of that column disappears.

Each time the operation is performed, a cost is incurred.
Since blocks of the same color have a property to draw each other together, the cost of the operation is the number of the pairs of blocks (p, q) such that:

The block p is in the selected column.

The two blocks p and q are horizontally adjacent (before pushing down the column).

The two blocks p and q have the same color.

Mr. Takahashi dismantles the object by repeating the operation H \times W times to get rid of all the blocks.
Compute the minimum total cost to dismantle the object.

Constraints

1 ≤ H ≤ 300

2 ≤ W ≤ 300

All characters c_{i,j} are lowercase English letters (a-z).

Partial Score

In test cases worth 300 points, W = 3.

Input

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

H W
c_{1,1}c_{1,2}..c_{1,W}
c_{2,1}c_{2,2}..c_{2,W}
:
c_{H,1}c_{H,2}..c_{H,W}

Output

Print the minimum total cost to dismantle the object.

Sample Input 1

2 3
rrr
brg

Sample Output 1

2

For example, the total cost of 2 can be achieved by performing the operation as follows and this is the minimum value.

Sample Input 2

6 3
xya
xya
ayz
ayz
xaz
xaz

Sample Output 2

0

The total cost of 0 can be achieved by first pushing down all blocks of the middle column, then all of the left column, and all of the right column.

Sample Input 3

4 2
ay
xa
xy
ay

Sample Output 3

0

The total cost of 0 can be achieved by the following operations:

pushing down the right column one row;

pushing down the left column one row;

pushing down all of the right column;

and pushing down all of the left column.

Sample Input 4

5 5
aaaaa
abbba
ababa
abbba
aaaaa

Sample Output 4

24

Sample Input 5

7 10
xxxxxxxxxx
ccccxxffff
cxxcxxfxxx
cxxxxxffff
cxxcxxfxxx
ccccxxfxxx
xxxxxxxxxx

Sample Output 5

130

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.


a, b, c = io.read("*n", "*n", "*n")
r = 0
if a ~= b and b ~= c and c ~= a then
	r = 3
elseif a == b and c ~= a then
	r = 2
elseif b == c and a ~= b then
	r = 2
elseif c == a and b ~= c then
	r = 2
else
	r = 1
end
print(r)

Score : 100 points

Problem Statement

AtCoDeer the deer recently bought three paint cans.
The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c.
Here, the color of each paint can is represented by an integer between 1 and 100, inclusive.

Since he is forgetful, he might have bought more than one paint can in the same color.
Count the number of different kinds of colors of these paint cans and tell him.

Constraints

1≦a,b,c≦100

Input

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

a b c

Output

Print the number of different kinds of colors of the paint cans.

Sample Input 1

3 1 4

Sample Output 1

3

Three different colors: 1, 3, and 4.

Sample Input 2

3 3 33

Sample Output 2

2

Two different colors: 3 and 33.


a, b = io.read("*n", "*n")
r = b
for i = 2, a do
  r = r * (b - 1)
end
print(r)

Score : 200 points

Problem Statement

There are N balls placed in a row.
AtCoDeer the deer is painting each of these in one of the K colors of his paint cans.
For aesthetic reasons, any two adjacent balls must be painted in different colors.

Find the number of the possible ways to paint the balls.

Constraints

1≦N≦1000

2≦K≦1000

The correct answer is at most 2^{31}-1.

Input

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

N K

Output

Print the number of the possible ways to paint the balls.

Sample Input 1

2 2

Sample Output 1

2

We will denote the colors by 0 and 1. There are two possible ways: we can either paint the left ball in color 0 and the right ball in color 1, or paint the left in color 1 and the right in color 0.

Sample Input 2

1 10

Sample Output 2

10

Since there is only one ball, we can use any of the ten colors to paint it. Thus, the answer is ten.


local n, k = io.read("*n", "*n")
local r = k
for i = 1, n - 1 do
  r = r * (k - 1)
end
print(r)


local w, h = io.read("*n", "*n")
local lim = (w + 1) * (h + 1) - 1
local cost = {}
local idxs = {}
for i = 1, w + h do
  cost[i] = io.read("*n")
  idxs[i] = i
end
table.sort(idxs, function(a, b) return cost[a] > cost[b] end)

local wuse, huse = 0, 0
local cnt = 0
local ret = 0LL
while cnt < lim do
  local idx = idxs[#idxs]
  table.remove(idxs)
  local u = 0
  if idx <= w then
    wuse = wuse + 1
    u = h + 1 - huse
  else
    huse = huse + 1
    u = w + 1 - wuse
  end
  cnt = cnt + u
  ret = ret + cost[idx] * u
end
local str = tostring(ret):gsub("LL", "")
print(str)

Score : 500 points

Problem Statement

On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.

There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.

The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.

Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.

Constraints

1 ≦ W,H ≦ 10^5

1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)

1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)

p_i (0 ≦ i ≦ W−1) is an integer.

q_j (0 ≦ j ≦ H−1) is an integer.

Input

Inputs are provided from Standard Input in the following form.

W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}

Output

Output an integer representing the minimum total cost.

Sample Input 1

2 2
3
5
2
7

Sample Output 1

29

It is enough to pave the following eight roads.

Road connecting houses at (0,0) and (0,1)

Road connecting houses at (0,1) and (1,1)

Road connecting houses at (0,2) and (1,2)

Road connecting houses at (1,0) and (1,1)

Road connecting houses at (1,0) and (2,0)

Road connecting houses at (1,1) and (1,2)

Road connecting houses at (1,2) and (2,2)

Road connecting houses at (2,0) and (2,1)

Sample Input 2

4 3
2
4
8
1
2
9
3

Sample Output 2

60


local n = io.read("*n")
local a = io.read("*n")
local ret = a - 1
local cur = 1
for i = 2, n do
  a = io.read("*n")
  if a <= cur then
    -- nothing
  elseif a == cur + 1 then
    cur = cur + 1
  else
    ret = ret + (a - 1) // (cur + 1)
  end
end
print(ret)

Score : 700 points

Problem Statement

N people are waiting in a single line in front of the Takahashi Store. The cash on hand of the i-th person from the front of the line is a positive integer A_i.

Mr. Takahashi, the shop owner, has decided on the following scheme: He picks a product, sets a positive integer P indicating its price, and shows this product to customers in order, starting from the front of the line. This step is repeated as described below.

At each step, when a product is shown to a customer, if price P is equal to or less than the cash held by that customer at the time, the customer buys the product and Mr. Takahashi ends the current step. That is, the cash held by the first customer in line with cash equal to or greater than P decreases by P, and the next step begins.

Mr. Takahashi can set the value of positive integer P independently at each step.

He would like to sell as many products as possible. However, if a customer were to end up with 0 cash on hand after a purchase, that person would not have the fare to go home. Customers not being able to go home would be a problem for Mr. Takahashi, so he does not want anyone to end up with 0 cash.

Help out Mr. Takahashi by writing a program that determines the maximum number of products he can sell, when the initial cash in possession of each customer is given.

Constraints

1 ≦ | N | ≦ 100000

1 ≦ A_i ≦ 10^9(1 ≦ i ≦ N)

All inputs are integers.

Input

Inputs are provided from Standard Inputs in the following form.

N
A_1
:
A_N

Output

Output an integer representing the maximum number of products Mr. Takahashi can sell.

Sample Input 1

3
3
2
5

Sample Output 1

3

As values of P, select in order 1, 4, 1.

Sample Input 2

15
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9

Sample Output 2

18


n, a, b = io.read("*n", "*n", "*n")
c = 0
for i = 1, n do
  z = io.read("*n")
  if z < a or b <= z then c = c + 1 end
end
print(c)

Score : 66 points

Problem Statement

Summer vacation ended at last and the second semester has begun.
You, a Kyoto University student, came to university and heard a rumor that somebody will barricade the entrance of your classroom.
The barricade will be built just before the start of the A-th class and removed by Kyoto University students just before the start of the B-th class.
All the classes conducted when the barricade is blocking the entrance will be cancelled and you will not be able to attend them.
Today you take N classes and class i is conducted in the t_i-th period.
You take at most one class in each period.
Find the number of classes you can attend.

Constraints

1 \leq N \leq 1000

1 \leq A < B \leq 10^9

1 \leq t_i \leq 10^9

All t_i values are distinct.

Input

N, A and B are given on the first line and t_i is given on the (i+1)-th line.

N A B
t1
:
tN

Output

Print the number of classes you can attend.

Sample Input 1

5 5 9
4
3
6
9
1

Sample Output 1

4

You can not only attend the third class.

Sample Input 2

5 4 9
5
6
7
8
9

Sample Output 2

1

You can only attend the fifth class.

Sample Input 3

4 3 6
9
6
8
1

Sample Output 3

4

You can attend all the classes.

Sample Input 4

2 1 2
1
2

Sample Output 4

1

You can not attend the first class, but can attend the second.


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, k = io.read("*n", "*n", "*l")
local t = {}
for i = 1, 26 do
  t[i] = 0
end
for i = 1, n do
  local s = io.read():byte(1) - 64
  t[s] = t[s] + 1
end
local avl = AvlTree.new(function(a, b) return t[a] > t[b] end, 1)
for i = 1, 26 do
  if 0 < t[i] then
    avl:push(i)
  end
end
local ret = 0
local z = {}
while true do
  local valid = true
  for i = 1, k do
    if avl:empty() then valid = false break end
    z[i] = avl:pop()
  end
  if not valid then break end
  ret = ret + 1
  for i = 1, k do
    t[z[i]] = t[z[i]] - 1
    if 0 < t[z[i]] then
      avl:push(z[i])
    end
  end
end
print(ret)

Score : 100 points

Problem Statement

Kyoto University Programming Contest is a programming contest voluntarily held by some Kyoto University students.
This contest is abbreviated as Kyoto University Programming Contest and called KUPC.

source: Kyoto University Programming Contest Information

The problem-preparing committee met to hold this year's KUPC and N problems were proposed there.
The problems are numbered from 1 to N and the name of i-th problem is P_i.
However, since they proposed too many problems, they decided to divide them into some sets for several contests.

They decided to divide this year's KUPC into several KUPCs by dividing problems under the following conditions.

One KUPC provides K problems.

Each problem appears at most once among all the KUPCs.

All the first letters of the problem names in one KUPC must be different.

You, one of the committee members, want to hold as many KUPCs as possible.
Write a program to find the maximum number of KUPCs that can be held this year.

Constraints

1 \leq N \leq 10^4

1 \leq K \leq 26

1 \leq |P_i| \leq 10

All characters in P_i are capital letters.

Note that, for each i and j (1 \leq i < j \leq N),
P_i \neq P_j are not necessarily satisfied.

Input

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

N K
P_1
:
P_N

Output

Print the maximum number of KUPCs that can be held on one line.

Sample Input 1

9 3
APPLE
ANT
ATCODER
BLOCK
BULL
BOSS
CAT
DOG
EGG

For example, three KUPCs can be held by dividing the problems as follows.

First: APPLE, BLOCK, CAT

Second: ANT, BULL, DOG

Third: ATCODER, BOSS, EGG

Sample Output 1

3

Sample Input 2

3 2
KU
KYOUDAI
KYOTOUNIV

Sample Output 2

0

No KUPC can be held.


local x=io.read()
local s,st=0,0
for i=1,#x do
    if x:sub(i,i)=="S" then
        s=s+1
    else
        if 0<s then
            s=s-1
            st=st+2
        end
    end
end
print(#x-st)

Score : 300 points

Problem Statement

We have a string X, which has an even number of characters. Half the characters are S, and the other half are T.

Takahashi, who hates the string ST, will perform the following operation 10^{10000} times:

Among the occurrences of ST in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.

Find the eventual length of X.

Constraints

2 ≦ |X| ≦ 200,000

The length of X is even.

Half the characters in X are S, and the other half are T.

Partial Scores

In test cases worth 200 points, |X| ≦ 200.

Input

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

X

Output

Print the eventual length of X.

Sample Input 1

TSTTSS

Sample Output 1

4

In the 1-st operation, the 2-nd and 3-rd characters of TSTTSS are removed.
X becomes TTSS, and since it does not contain ST anymore, nothing is done in the remaining 10^{10000}-1 operations.
Thus, the answer is 4.

Sample Input 2

SSTTST

Sample Output 2

0

X will eventually become an empty string: SSTTST ⇒ STST ⇒ ST ⇒ ``.

Sample Input 3

TSSTTTSS

Sample Output 3

4

X will become: TSSTTTSS ⇒ TSTTSS ⇒ TTSS.


local x=io.read()
local y=0
for k,v in x:gmatch("%bST") do
    y=y+#k
end
print(#x-#y)


local n = io.read()

local function getABC(str)

	local result = 0
	local function getStr(ss)
		local p = string.find(ss, "ST")
		if not p then
			result = #ss
			return
		end
		local s1 = string.sub(ss, 1, p-1)
		local s2 = string.sub(ss, p+2, #ss)
		local s = s1 .. s2
		getStr(s)
	end
	getStr(str)

	if result % 2 ~= 0 then
		result = result + 1
	end
	return result
end

local r = getABC(n)
print(r)


local mfl, mce = math.floor, math.ceil
local mmi, mma = math.min, math.max
local bls, brs = bit.lshift, bit.rshift
-- local pow2 = {1}
-- for i = 2, 26 do
--   pow2[i] = 2 * pow2[i - 1]
-- end
-- local function bls(a, b)
--   return a * pow2[b + 1]
-- end
-- local function brs(a, b)
--   return a // pow2[b + 1]
-- end

local SegTree = {}
SegTree.updateAll = function(self)
  for i = self.stagenum - 1, 1, -1 do
    local cnt = bls(1, i - 1)
    for j = 1, cnt do
      self.stage[i][j] = self.func(self.stage[i + 1][j * 2 - 1], self.stage[i + 1][j * 2])
    end
  end
end
SegTree.create = function(self, n, func, emptyvalue)
  self.func, self.emptyvalue = func, emptyvalue
  local stagenum, mul = 1, 1
  self.stage = {{}}
  while mul < n do
    mul, stagenum = mul * 2, stagenum + 1
    self.stage[stagenum] = {}
  end
  self.stagenum = stagenum
  self.left_stage = {}
  for i = 1, n do
    local sp, sz = 1, bls(1, stagenum - 1)
    while(i - 1) % sz ~= 0 do
      sp, sz = sp + 1, brs(sz, 1)
    end
    self.left_stage[i] = sp
  end
  self.sz_stage = {}
  local tmp, sp = 1, stagenum
  for i = 1, n do
    if tmp * 2 == i then tmp, sp = tmp * 2, sp - 1 end
    self.sz_stage[i] = sp
  end
  for i = 1, mul do self.stage[stagenum][i] = emptyvalue end
  self:updateAll()
end
SegTree.setValue = function(self, idx, value, silent)
  self.stage[self.stagenum][idx] = value
  if not silent then
    for i = self.stagenum - 1, 1, -1 do
      local dst = brs(idx + 1, 1)
      local rem = dst * 4 - 1 - idx
      self.stage[i][dst] = self.func(self.stage[i + 1][idx], self.stage[i + 1][rem])
      idx = dst
    end
  end
end
SegTree.right_bound = function(self, val, left, right)
  local ret, retpos = self.emptyvalue, left - 1
  local l, r = left, right
  local stage = mma(self.left_stage[left], self.sz_stage[right - left + 1])
  local stagenum = self.stagenum
  while true do
    local sz = bls(1, stagenum - stage)
    local tmp = self.func(ret, self.stage[stage][mce(l / sz)])
    if tmp < val then
      ret, retpos = tmp, l + sz - 1
      if retpos == right then break end
      if l + sz <= r then
        l = l + sz
        stage = mma(self.left_stage[l], self.sz_stage[r - l + 1])
      else break end
    else
      if sz ~= 1 then stage, r = stage + 1, l + sz - 2
      else break end
    end
  end
  return retpos + 1
end
SegTree.left_bound = function(self, val, left, right)
  local ret, retpos = self.emptyvalue, right + 1
  local stage, l, r = 1, left, right
  local stagenum = self.stagenum
  while true do
    local sz = bls(1, stagenum - stage)
    while r % sz ~= 0 or r + 1 - l < sz do
      stage = stage + 1
      sz = bls(1, stagenum - stage)
    end
    local tmp = self.func(ret, self.stage[stage][mfl(r / sz)])
    if tmp < val then
      ret, retpos = tmp, r - sz + 1
      if l + sz <= r then stage, l, r = 1, l, r - sz
      else break end
    else
      if sz ~= 1 then stage, l, r = stage + 1, r - sz + 2, r
      else break end
    end
  end
  return retpos - 1
end
SegTree.new = function(n, func, emptyvalue)
  local obj = {}
  setmetatable(obj, {__index = SegTree})
  obj:create(n, func, emptyvalue)
  return obj
end

local n = io.read("*n")
local st = SegTree.new(n, mma, -n-1)
local invpos = {}
for i = 1, n do invpos[i] = i end
for i = 1, n do
  local z = io.read("*n")
  st:setValue(i, -z, true)
  invpos[z] = i
end
st:updateAll()
local ret = 0LL
for i = 1, n do
  local pos = invpos[i]
  local rb = st:right_bound(-i + 1, pos, n)
  local lb = st:left_bound(-i + 1, 1, pos)
  ret = ret + 1LL * i * (rb - pos) * (pos - lb)
end
local str = tostring(ret):gsub("LL", "")
print(str)

Score : 400 points

Problem Statement

One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.

Find the following:

Constraints

1 ≦ N ≦ 200,000

(a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).

Input

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

N
a_1 a_2 ... a_N

Output

Print the answer.

Note that the answer may not fit into a 32-bit integer.

Sample Input 1

3
2 1 3

Sample Output 1

9

Sample Input 2

4
1 3 2 4

Sample Output 2

19

Sample Input 3

8
5 4 8 1 2 6 7 3

Sample Output 3

85


local s = io.read()
print(s:sub(1, 4) .. " " .. s:sub(5, 12))

Score : 100 points

Problem Statement

This contest is CODE FESTIVAL.
However, Mr. Takahashi always writes it CODEFESTIVAL, omitting the single space between CODE and FESTIVAL.

So he has decided to make a program that puts the single space he omitted.

You are given a string s with 12 letters.
Output the string putting a single space between the first 4 letters and last 8 letters in the string s.

Constraints

s contains exactly 12 letters.

All letters in s are uppercase English letters.

Input

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

s

Output

Print the string putting a single space between the first 4 letters and last 8 letters in the string s.
Put a line break at the end.

Sample Input 1

CODEFESTIVAL

Sample Output 1

CODE FESTIVAL

Putting a single space between the first 4 letters and last 8 letters in CODEFESTIVAL makes it CODE FESTIVAL.

Sample Input 2

POSTGRADUATE

Sample Output 2

POST GRADUATE

Sample Input 3

ABCDEFGHIJKL

Sample Output 3

ABCD EFGHIJKL


local s=io.read()
local sable={}
for i=1,#s do
    sable[i]=s:byte(i)-96
end
local k=io.read("n")
for i=1,#s do
    local a=sable[i]
    if k>=27-a and a>1 then
        sable[i]=string.char(97)
        k=k-(27-a)
    elseif i==#s then
        sable[i]=string.char((sable[i]+k)%27+96)
    else
        sable[i]=string.char(sable[i]+96)
    end
end
print(table.concat(sable,""))

Score : 400 points

Problem Statement

Mr. Takahashi has a string s consisting of lowercase English letters.
He repeats the following operation on s exactly K times.

Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of z is a.

For example, if you perform an operation for the second letter on aaz, aaz becomes abz.
If you then perform an operation for the third letter on abz, abz becomes aba.

Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s.
Find the such string.

Constraints

1≤|s|≤10^5

All letters in s are lowercase English letters.

1≤K≤10^9

Input

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

s
K

Output

Print the lexicographically smallest string after performing exactly K operations on s.

Sample Input 1

xyz
4

Sample Output 1

aya

For example, you can perform the following operations: xyz, yyz, zyz, ayz, aya.

Sample Input 2

a
25

Sample Output 2

z

You have to perform exactly K operations.

Sample Input 3

codefestival
100

Sample Output 3

aaaafeaaivap


Score : 100 points

Problem Statement

You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively.

An example of a trapezoid

Find the area of this trapezoid.

Constraints

1≦a≦100

1≦b≦100

1≦h≦100

All input values are integers.

h is even.

Input

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

a
b
h

Output

Print the area of the given trapezoid. It is guaranteed that the area is an integer.

Sample Input 1

3
4
2

Sample Output 1

7

When the lengths of the upper base, lower base, and height are 3, 4, and 2, respectively, the area of the trapezoid is (3+4)×2/2 = 7.

Sample Input 2

4
4
4

Sample Output 2

16

In this case, a parallelogram is given, which is also a trapezoid.


local S={}
S["a"]=io.read()
S["b"]=io.read()
S["c"]=io.read()

local discard="a"
while S["a"]~="" and S["b"]~="" and S["c"]~="" do
    discard=S[discard]:sub(1,1)
    S[discard]=S[discard]:sub(2,#S[discard])
    if S[discard]=="" then
        print(discard:upper())
        break
    end
end

Score : 200 points

Problem Statement

Alice, Bob and Charlie are playing Card Game for Three, as below:

At first, each of the three players has a deck consisting of some number of cards. Each card has a letter a, b or c written on it. The orders of the cards in the decks cannot be rearranged.

The players take turns. Alice goes first.

If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says a, Alice takes the next turn.)

If the current player's deck is empty, the game ends and the current player wins the game.

You are given the initial decks of the players.
More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way.

Determine the winner of the game.

Constraints

1≦|S_A|≦100

1≦|S_B|≦100

1≦|S_C|≦100

Each letter in S_A, S_B, S_C is a, b or c.

Input

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

S_A
S_B
S_C

Output

If Alice will win, print A. If Bob will win, print B. If Charlie will win, print C.

Sample Input 1

aca
accc
ca

Sample Output 1

A

The game will progress as below:

Alice discards the top card in her deck, a. Alice takes the next turn.

Alice discards the top card in her deck, c. Charlie takes the next turn.

Charlie discards the top card in his deck, c. Charlie takes the next turn.

Charlie discards the top card in his deck, a. Alice takes the next turn.

Alice discards the top card in her deck, a. Alice takes the next turn.

Alice's deck is empty. The game ends and Alice wins the game.

Sample Input 2

abcb
aacb
bccc

Sample Output 2

C


sa = io.read()
sb = io.read()
sc = io.read()
local ta, tb, tc = 1, 1, 1
cur = "a"
while true do
  if cur == "a" then
    if #sa < ta then
      print("A") os.exit()
    end
    cur = sa:sub(ta, ta)
    ta = ta + 1
  elseif cur == "b" then
    if #sb < tb then
      print("B") os.exit()
    end
    cur = sb:sub(tb, tb)
    tb = tb + 1
  else
    if #sc < tc then
      print("C") os.exit()
    end
    cur = sc:sub(tc, tc)
    tc = tc + 1
  end
end


local tgt = 1
local strs = {}
local pos = {1, 1, 1}
local tmp = ""
for i = 1, 3 do strs[i] = io.read() end
while pos[tgt] <= #(strs[tgt]) do
  tmp = strs[tgt]:sub(pos[tgt], pos[tgt])
  pos[tgt] = pos[tgt] + 1
  tgt = string.byte(tmp) - 96
end
print(string.char(tgt + 64))


local ior = io.read
local str = ior()
local ppos = #str - 1
local tot = (1<<ppos) - 1
local sum = 0
for i = 0, tot do
  local tsum = 0
  local t = 0
  local it = i
  for j = 1, ppos do
    if(it % 2 == 1) then
      t = t * 10 + tonumber(str:sub(j, j))
      tsum = tsum + t
      t = 0
    else
      t = t * 10 + tonumber(str:sub(j, j))
    end
    it = it // 2
  end
  tsum = tsum + t * 10 + tonumber(str:sub(#str, #str))
  sum = sum + tsum
end
print(sum)

Score : 300 points

Problem Statement

You are given a string S consisting of digits between 1 and 9, inclusive.
You can insert the letter + into some of the positions (possibly none) between two letters in this string.
Here, + must not occur consecutively after insertion.

All strings that can be obtained in this way can be evaluated as formulas.

Evaluate all possible formulas, and print the sum of the results.

Constraints

1 \leq |S| \leq 10

All letters in S are digits between 1 and 9, inclusive.

Input

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

S

Output

Print the sum of the evaluated value over all possible formulas.

Sample Input 1

125

Sample Output 1

176

There are 4 formulas that can be obtained: 125, 1+25, 12+5 and 1+2+5. When each formula is evaluated,

125

1+25=26

12+5=17

1+2+5=8

Thus, the sum is 125+26+17+8=176.

Sample Input 2

9999999999

Sample Output 2

12656242944


local h, w, n = io.read("*n", "*n", "*n")
local t = {}

local function add(i_h, i_w)
  local idx = i_h .. ":" .. i_w --(i_h - 1) * 1000000000 + i_w
  local v = t[idx]
  if v then t[idx] = v + 1 else t[idx] = 1 end
end

local function subtask(ah, bw)
  if 3 <= bw then add(ah, bw - 2) end
  if 2 <= bw and bw < w then add(ah, bw - 1) end
  if bw < w - 1 then add(ah, bw) end
end

for i = 1, n do
  local ah, bw = io.read("*n", "*n")
  if 3 <= ah then subtask(ah - 2, bw) end
  if 2 <= ah and ah  < h then subtask(ah - 1, bw) end
  if ah < h - 1 then subtask(ah, bw) end
end

local sum = 0
local tbl = {}
for i = 1, 9 do tbl[i] = 0 end
for k, v in pairs(t) do
  tbl[v] = tbl[v] + 1
  sum = sum + 1
end
local rem = (h - 2LL) * (w - 2LL) - sum
local str = tostring(rem):gsub("LL", "")
print(str)
for i = 1, 9 do print(tbl[i]) end

Score : 400 points

Problem Statement

We have a grid with H rows and W columns. At first, all cells were painted white.

Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column.

Compute the following:

For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?

Constraints

3 \leq H \leq 10^9

3 \leq W \leq 10^9

0 \leq N \leq min(10^5,H×W)

1 \leq a_i \leq H (1 \leq i \leq N)

1 \leq b_i \leq W (1 \leq i \leq N)

(a_i, b_i) \neq (a_j, b_j) (i \neq j)

Input

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

H W N
a_1 b_1
:
a_N b_N

Output

Print 10 lines.
The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.

Sample Input 1

4 5 8
1 1
1 4
1 5
2 3
3 1
3 2
3 4
4 4

Sample Output 1

0
0
0
2
4
0
0
0
0
0

There are six subrectangles of size 3×3. Two of them contain three black cells each, and the remaining four contain four black cells each.

Sample Input 2

10 10 20
1 1
1 4
1 9
2 5
3 10
4 2
4 7
5 9
6 4
6 6
6 7
7 1
7 3
7 7
8 1
8 5
8 10
9 2
10 4
10 9

Sample Output 2

4
26
22
10
2
0
0
0
0
0

Sample Input 3

1000000000 1000000000 0

Sample Output 3

999999996000000004
0
0
0
0
0
0
0
0
0


local mma = math.max
local mfl, mce, mmi = math.floor, math.ceil, math.min
local SegForAvl = {}
SegForAvl.updateAll = function(self)
  for i = self.stagenum - 1, 1, -1 do
    for j = 1, self.cnt[i] do
      self.stage[i][j] = self.stage[i + 1][j * 2 - 1]
    end
  end
end
SegForAvl.create = function(self, n)
  local stagenum, mul = 1, 1
  self.cnt, self.stage, self.size = {1}, {{}}, {}
  while mul < n do
    mul, stagenum = mul * 2, stagenum + 1
    self.cnt[stagenum], self.stage[stagenum] = mul, {}
  end
  for i = 1, stagenum do self.size[i] = self.cnt[stagenum + 1 - i] end
  self.stagenum = stagenum
  for i = 1, mul do self.stage[stagenum][i] = i end
  self:updateAll()
end
SegForAvl.hold = function(self)
  local idx = self.stage[1][1]
  local ridx = idx
  self.stage[self.stagenum][idx] = false
  for i = self.stagenum - 1, 1, -1 do
    local dst = mce(idx / 2)
    local rem = dst * 4 - 1 - idx
    self.stage[i][dst] = self.stage[i + 1][idx] or self.stage[i + 1][rem]
    idx = dst
  end
  return ridx
end
SegForAvl.release = function(self, idx)
  self.stage[self.stagenum][idx] = idx
  for i = self.stagenum - 1, 1, -1 do
    local dst = mce(idx / 2)
    local rem = dst * 4 - 1 - idx
    self.stage[i][dst] = self.stage[i + 1][idx] or self.stage[i + 1][rem]
    idx = dst
  end
end
SegForAvl.new = function(n)
  local obj = {}
  setmetatable(obj, {__index = SegForAvl})
  obj:create(n)
  return obj
end

local SegAvlTree = {}
SegAvlTree.makenode = function(self, val, parent)
  local i = self.seg:hold()
  self.v[i], self.u[i], self.p[i] = val, true, parent
  self.lc[i], self.rc[i], self.l[i], self.r[i] = 0, 0, 0, 0
  return i
end
SegAvlTree.create = function(self, lessthan, n)
  self.lessthan = lessthan
  self.seg = SegForAvl.new(n)
  -- value, leftCount, rightCount, left, right, parent, used
  self.v, self.lc, self.rc, self.l, self.r, self.p, self.u = {}, {}, {}, {}, {}, {}, {}
  for i = 1, n do
    self.v[i], self.u[i], self.p[i] = 0, false, 0
    self.lc[i], self.rc[i], self.l[i], self.r[i] = 0, 0, 0, 0
  end
end

SegAvlTree.recalcCount = function(self, i)
  if self.u[i] then
    if self.u[self.l[i]] then self.lc[i] = 1 + mma(self.lc[self.l[i]], self.rc[self.l[i]])
    else self.lc[i] = 0
    end
    if self.u[self.r[i]] then self.rc[i] = 1 + mma(self.lc[self.r[i]], self.rc[self.r[i]])
    else self.rc[i] = 0
    end
  end
end
SegAvlTree.recalcCountAll = function(self, i)
  while self.u[i] do
    self:recalcCount(i)
    i = self.p[i]
  end
end

SegAvlTree.rotR = function(self, child, parent)
  local granp = self.p[parent]
  self.r[child], self.l[parent] = parent, self.r[child]
  self.p[child], self.p[parent] = granp, child
  if self.u[self.l[parent]] then self.p[self.l[parent]] = parent end
  if self.u[granp] then
    if self.l[granp] == parent then
      self.l[granp] = child
    else
      self.r[granp] = child
    end
  else
    self.root = child
  end
  self:recalcCountAll(parent)
end

SegAvlTree.rotL = function(self, child, parent)
  local granp = self.p[parent]
  self.l[child], self.r[parent] = parent, self.l[child]
  self.p[child], self.p[parent] = granp, child
  if self.u[self.r[parent]] then self.p[self.r[parent]] = parent end
  if self.u[granp] then
    if self.r[granp] == parent then
      self.r[granp] = child
    else
      self.l[granp] = child
    end
  else
    self.root = child
  end
  self:recalcCountAll(parent)
end

SegAvlTree.add = function(self, val)
  if not self.u[self.root] then self.root = self:makenode(val, 0) return end
  local pos = self.root
  local added = false
  while not added do
    if self.lessthan(val, self.v[pos]) then
      if self.u[self.l[pos]] then
        pos = self.l[pos]
      else
        self.l[pos] = self:makenode(val, pos)
        pos = self.l[pos]
        added = true
      end
    else
      if self.u[self.r[pos]] then
        pos = self.r[pos]
      else
        self.r[pos] = self:makenode(val, pos)
        pos = self.r[pos]
        added = true
      end
    end
  end
  while self.u[pos] do
    local child, parent = pos, self.p[pos]
    if not self.u[parent] then
      break
    end
    self:recalcCount(parent)
    if self.l[parent] == child then
      if self.lc[parent] - 1 == self.rc[parent] then
        pos = parent
      elseif self.lc[parent] - 2 == self.rc[parent] then
        self:recalcCount(child)
        if self.lc[child] - 1 == self.rc[child] then
          self:rotR(child, parent)
        else
          local cr = self.r[child]
          self:rotL(cr, child)
          self:rotR(cr, parent)
        end
        pos = nil
      else
        self:recalcCountAll(child)
        pos = nil
      end
    else -- parent.r == child
      if self.rc[parent] - 1 == self.lc[parent] then
        pos = parent
      elseif self.rc[parent] - 2 == self.lc[parent] then
        self:recalcCount(child)
        if self.rc[child] - 1 == self.lc[child] then
          self:rotL(child, parent)
        else
          local cl = self.l[child]
          self:rotR(cl, child)
          self:rotL(cl, parent)
        end
        pos = nil
      else
        self:recalcCountAll(child)
        pos = nil
      end
    end
  end
end

SegAvlTree.rmsub = function(self, node)
  while self.u[node] do
    self:recalcCount(node)
    if self.lc[node] == self.rc[node] then
      node = self.p[node]
    elseif self.lc[node] + 1 == self.rc[node] then
      self:recalcCountAll(self.p[node])
      node = nil
    else
      if self.lc[self.r[node]] == self.rc[self.r[node]] then
        self:rotL(self.r[node], node)
        node = nil
      elseif self.lc[self.r[node]] + 1 == self.rc[self.r[node]] then
        local nr = self.r[node]
        self:rotL(nr, node)
        node = nr
      else
        local nrl = self.l[self.r[node]]
        self:rotR(nrl, self.r[node])
        self:rotL(nrl, node)
        node = nrl
      end
    end
  end
end

SegAvlTree.pop = function(self)
  local node = self.root
  while self.u[self.l[node]] do
    node = self.l[node]
  end
  local v = self.v[node]
  if self.u[self.p[node]] then
    if self.u[self.r[node]] then self.p[self.r[node]] = self.p[node] end
    self.l[self.p[node]] = self.r[node]
    self:rmsub(self.p[node])
  else
    if self.r[node] then self.p[self.r[node]] = 0 end
    self.root = self.r[node]
  end
  self.seg:release(node)
  return v
end

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

local n, m = io.read("*n", "*n")
local ps, qs, cs = {}, {}, {}
local companies = {}
local stations = {}
for i = 1, n do
  stations[i] = {}
end
for i = 1, m do
  local p, q, c = io.read("*n", "*n", "*n")
  if not companies[c] then companies[c] = {} end
  if stations[p][c] and stations[q][c] then
    if stations[p][c] ~= stations[q][c] then
      local ci_p, ci_q = stations[p][c], stations[q][c]
      if #companies[c][ci_p] < #companies[c][ci_q] then
        for j = 1, #companies[c][ci_p] do
          local st = companies[c][ci_p][j]
          stations[st][c] = ci_q
          table.insert(companies[c][ci_q], st)
        end
        companies[c][ci_p] = {}
      else
        for j = 1, #companies[c][ci_q] do
          local st = companies[c][ci_q][j]
          stations[st][c] = ci_p
          table.insert(companies[c][ci_p], st)
        end
        companies[c][ci_q] = {}
      end
    end
  elseif not stations[p][c] and not stations[q][c] then
    table.insert(companies[c], {p, q})
    stations[p][c] = #companies[c]
    stations[q][c] = #companies[c]
  elseif stations[p][c] then
    local idx = stations[p][c]
    stations[q][c] = idx
    table.insert(companies[c][idx], q)
  else
    local idx = stations[q][c]
    stations[p][c] = idx
    table.insert(companies[c][idx], p)
  end
end
local len = {}
local asked = {}
local tasknum = 0
local done = 0
local tasktree = SegAvlTree.new(function(x, y) return len[x[1]][x[2]] < len[y[1]][y[2]] end, 200000)
for i = 1, n do
  asked[i] = false
  len[i] = {}
  for company, group in pairs(stations[i]) do
    len[i][company] = -1
  end
end
local function addtask(station, company)
  tasknum = tasknum + 1
  tasktree:add({station, company})
end
for company, group in pairs(stations[1]) do
  len[1][company] = 1
  addtask(1, company)
end
local e,f = 0,0
while done < tasknum do
  done = done + 1
  local taskidx = tasktree:pop()
  local src_station, src_company = taskidx[1], taskidx[2]
  local src_group = stations[src_station][src_company]
  local srclen = len[src_station][src_company]
  if not asked[src_station] then
    asked[src_station] = true
    for dst_company, _g in pairs(stations[src_station]) do
      if src_company ~= dst_company then
        local dstlen = len[src_station][dst_company]
        if dstlen == -1 or srclen + 1 < dstlen then
          len[src_station][dst_company] = srclen + 1
          addtask(src_station, dst_company)
        end
      end
    end
  end
  if 0 < #companies[src_company][src_group] then
    for i = 1, #companies[src_company][src_group] do
      local dst_station = companies[src_company][src_group][i]
      local dstlen = len[dst_station][src_company]
      if dstlen == -1 or srclen < dstlen then
        len[dst_station][src_company] = srclen
        addtask(dst_station, src_company)
      end
    end
    companies[src_company][src_group] = {}
  end
end
local ret = -1
local mmi = math.min
for company, group in pairs(stations[n]) do
  local dstlen = len[n][company]
  if 0 < dstlen then
    if ret == -1 then
      ret = dstlen
    else
      ret = mmi(ret, dstlen)
    end
  end
end
print(ret)

Score : 600 points

Problem Statement

Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.

The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.

You can change trains at a station where multiple lines are available.

The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.

Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq 2×10^5

1 \leq p_i \leq N (1 \leq i \leq M)

1 \leq q_i \leq N (1 \leq i \leq M)

1 \leq c_i \leq 10^6 (1 \leq i \leq M)

p_i \neq q_i (1 \leq i \leq M)

Input

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

N M
p_1 q_1 c_1
:
p_M q_M c_M

Output

Print the minimum required fare. If it is impossible to get to station N by subway, print -1 instead.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1


local n, m = io.read("*n", "*n")
local ps, qs, cs = {}, {}, {}
local companies = {}
local stations = {}
for i = 1, n do
  stations[i] = {}
end
local compmax = 1000000
for i = 1, compmax do
  companies[i] = {}
end
for i = 1, m do
  local p, q, c = io.read("*n", "*n", "*n")
  if stations[p][c] and stations[q][c] then
    if stations[p][c] ~= stations[q][c] then
      local ci_p, ci_q = stations[p][c], stations[q][c]
      if #companies[c][ci_p] < #companies[c][ci_q] then
        for j = 1, #companies[c][ci_p] do
          local st = companies[c][ci_p][j]
          stations[st][c] = ci_q
          table.insert(companies[c][ci_q], st)
        end
        companies[c][ci_p] = {}
      else
        for j = 1, #companies[c][ci_q] do
          local st = companies[c][ci_q][j]
          stations[st][c] = ci_p
          table.insert(companies[c][ci_p], st)
        end
        companies[c][ci_q] = {}
      end
    end
  elseif not stations[p][c] and not stations[q][c] then
    table.insert(companies[c], {p, q})
    stations[p][c] = #companies[c]
    stations[q][c] = #companies[c]
  elseif stations[p][c] then
    local idx = stations[p][c]
    stations[q][c] = idx
    table.insert(companies[c][idx], q)
  else
    local idx = stations[q][c]
    stations[p][c] = idx
    table.insert(companies[c][idx], p)
  end
end
local len = {}
local taskstate = {}
local tasks = {}
local tasknum = 0
local done = 0
local tasklim = 200000
for i = 1, compmax do
  len[i] = {}
  taskstate[i] = {}
  for j = 1, #companies[i] do
    len[i][j] = -1
    taskstate[i][j] = false
  end
end
local function addtask(company, group)
  if not taskstate[company][group] then
    taskstate[company][group] = true
    tasknum = tasknum + 1
    local taskidx = tasknum % tasklim
    if taskidx == 0 then taskidx = tasklim end
    tasks[taskidx] = {company, group}
  end
end
for company, group in pairs(stations[1]) do
  len[company][group] = 1
  addtask(company, group)
end
local e = 0
while done < tasknum do
  done = done + 1
  local taskidx = done % tasklim
  if taskidx == 0 then taskidx = tasklim end
  local src_company, src_group = tasks[taskidx][1], tasks[taskidx][2]
  local srclen = len[src_company][src_group]
  taskstate[src_company][src_group] = false
  for i = 1, #companies[src_company][src_group] do
    local station = companies[src_company][src_group][i]
    for dst_company, dst_group in pairs(stations[station]) do
      e = e + 1
      if src_company ~= dst_company then
        local dstlen = len[dst_company][dst_group]
        if dstlen == -1 or srclen + 1 < dstlen then
          len[dst_company][dst_group] = srclen + 1
          addtask(dst_company, dst_group)
        end
      end
    end
  end
end
local ret = -1
local mmi = math.min
for company, group in pairs(stations[n]) do
  local dstlen = len[company][group]
  if 0 < dstlen then
    if ret == -1 then
      ret = dstlen
    else
      ret = mmi(ret, dstlen)
    end
  end
end
print(ret)


a={}
for i=1,3 do
    table.insert(a,io.read("*n"))
end
table.sort(a)
print(a[1]*a[2]*(a[3]%2))

Score : 200 points

Problem Statement

We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:

There is at least one red block and at least one blue block.

The union of all red blocks forms a rectangular parallelepiped.

The union of all blue blocks forms a rectangular parallelepiped.

Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.

Constraints

2≤A,B,C≤10^9

Input

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

A B C

Output

Print the minimum possible difference between the number of red blocks and the number of blue blocks.

Sample Input 1

3 3 3

Sample Output 1

9

For example, Snuke can paint the blocks as shown in the diagram below.
There are 9 red blocks and 18 blue blocks, thus the difference is 9.

Sample Input 2

2 2 4

Sample Output 2

0

For example, Snuke can paint the blocks as shown in the diagram below.
There are 8 red blocks and 8 blue blocks, thus the difference is 0.

Sample Input 3

5 3 5

Sample Output 3

15

For example, Snuke can paint the blocks as shown in the diagram below.
There are 45 red blocks and 30 blue blocks, thus the difference is 9.


local n = io.read('*n')
local x = io.read("*n")
io.read("*l")
local a = {}
for i = 1, n do
	a[i] = io.read("*n")
end
local b = {}
for i = 1, n do
	b[i] = {}
	b[i][0] = a[i]
	for j = 1, n - 1 do
		local idx = i - j
		if idx <= 0 then
			idx = idx + n
		end
		b[i][j] = math.min(a[idx], b[i][j - 1])
	end
end

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

Score : 400 points

Problem Statement

Snuke lives in another world, where slimes are real creatures and kept by some people.
Slimes come in N colors. Those colors are conveniently numbered 1 through N.
Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him a_i seconds.

Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N-1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.

Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

2≤N≤2,000

a_i are integers.

1≤a_i≤10^9

x is an integer.

1≤x≤10^9

Input

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

N x
a_1 a_2 ... a_N

Output

Find the minimum time that Snuke needs to have slimes in all N colors.

Sample Input 1

2 10
1 100

Sample Output 1

12

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.

Catch a slime in color 1. This takes 1 second.

Sample Input 2

3 10
100 1 100

Sample Output 2

23

Snuke can act as follows:

Catch a slime in color 2. This takes 1 second.

Cast the spell. The color of the slime changes: 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Cast the soell. The color of each slime changes: 3 → 1, 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Sample Input 3

4 10
1 2 3 4

Sample Output 3

10

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Catch a slime in color 2. This takes 2 seconds.

Catch a slime in color 3. This takes 3 seconds.

Catch a slime in color 4. This takes 4 seconds.


local mma = math.max
local n, k = io.read("*n", "*n")
local edge = {}
local len = {}
local childcnt = {}
local parent = {}
for i = 1, n do
  edge[i] = {}
  len[i] = -1
  childcnt[i] = 0
  parent[i] = 0
end
len[1] = 0
local ret = io.read("*n") == 1 and 0 or 1
for i = 2, n do
  local a = io.read("*n")
  table.insert(edge[a], i)
  table.insert(edge[i], a)
end
local tasks = {1}
for it = 1, n do
  local src = tasks[it]
  for i = 1, #edge[src] do
    local dst = edge[src][i]
    if len[dst] == -1 then
      len[dst] = len[src] + 1
      table.insert(tasks, dst)
      childcnt[src] = childcnt[src] + 1
      parent[dst] = src
    end
  end
end
local maxlen = {}
tasks = {}
for i = 1, n do
  maxlen[i] = len[i]
  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]
  if k < maxlen[c] then
    if maxlen[c] - len[c] == k - 1 then
      ret = ret + 1
    else
      maxlen[p] = mma(maxlen[p], maxlen[c])
    end
  else
    maxlen[p] = mma(maxlen[p], maxlen[c])
  end
  childcnt[p] = childcnt[p] - 1
  if childcnt[p] == 0 then
    table.insert(tasks, p)
  end
end
print(ret)

Score : 800 points

Problem Statement

There are N towns in Snuke Kingdom, conveniently numbered 1 through N.
Town 1 is the capital.

Each town in the kingdom has a Teleporter, a facility that instantly transports a person to another place.
The destination of the Teleporter of town i is town a_i (1≤a_i≤N).
It is guaranteed that one can get to the capital from any town by using the Teleporters some number of times.

King Snuke loves the integer K.
The selfish king wants to change the destination of the Teleporters so that the following holds:

Starting from any town, one will be at the capital after using the Teleporters exactly K times in total.

Find the minimum number of the Teleporters whose destinations need to be changed in order to satisfy the king's desire.

Constraints

2≤N≤10^5

1≤a_i≤N

One can get to the capital from any town by using the Teleporters some number of times.

1≤K≤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 minimum number of the Teleporters whose destinations need to be changed in order to satisfy King Snuke's desire.

Sample Input 1

3 1
2 3 1

Sample Output 1

2

Change the destinations of the Teleporters to a = (1,1,1).

Sample Input 2

4 2
1 1 2 2

Sample Output 2

0

There is no need to change the destinations of the Teleporters, since the king's desire is already satisfied.

Sample Input 3

8 2
4 1 2 3 1 2 3 4

Sample Output 3

3

For example, change the destinations of the Teleporters to a = (1,1,2,1,1,2,2,4).


a, b = io.read("*n", "*n")
c, d = io.read("*n", "*n")
if a <= b then
  print(c * a)
else
  print(c * b + d * (a - b))
end

Score : 100 points

Problem Statement

There is a hotel with the following accommodation fee:

X yen (the currency of Japan) per night, for the first K nights

Y yen per night, for the (K+1)-th and subsequent nights

Tak is staying at this hotel for N consecutive nights.
Find his total accommodation fee.

Constraints

1 \leq N, K \leq 10000

1 \leq Y < X \leq 10000

N,\,K,\,X,\,Y are integers.

Input

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

N
K
X
Y

Output

Print Tak's total accommodation fee.

Sample Input 1

5
3
10000
9000

Sample Output 1

48000

The accommodation fee is as follows:

10000 yen for the 1-st night

10000 yen for the 2-nd night

10000 yen for the 3-rd night

9000 yen for the 4-th night

9000 yen for the 5-th night

Thus, the total is 48000 yen.

Sample Input 2

2
3
10000
9000

Sample Output 2

20000


local mfl = math.floor
local mce, msq = math.ceil, math.sqrt
local function getprimes(x)
  local primes = {}
  local allnums = {}
  for i = 1, x do allnums[i] = true end
  for i = 2, x do
    if allnums[i] then
      table.insert(primes, i)
      local lim = x // i
      for j = 2, lim do
        allnums[j * i] = false
      end
    end
  end
  return primes
end

local function getdivisorparts(x, primes)
  local prime_num = #primes
  local tmp = {}
  local lim = mce(msq(x))
  local primepos = 1
  local dv = primes[primepos]
  while primepos <= prime_num and dv <= lim do
    if x % dv == 0 then
      local asdf = {}
      asdf.p = dv
      asdf.cnt = 1
      x = x // dv
      while x % dv == 0 do
        x = x // dv
        asdf.cnt = asdf.cnt + 1
      end
      table.insert(tmp, asdf)
      lim = mce(msq(x))
    end
    if primepos == prime_num then break end
    primepos = primepos + 1
    dv = primes[primepos]
  end
  if x ~= 1 then
    local asdf = {}
    asdf.p, asdf.cnt = x, 1
    table.insert(tmp, asdf)
  end
  return tmp
end

local function getdivisor(divisorparts)
  local t = {}
  local pat = 1
  local len = #divisorparts
  local allpat = 1
  for i = 1, len do
    allpat = allpat * (1 + divisorparts[i].cnt)
  end
  for t_i_pat = 0, allpat - 1 do
    local div = allpat
    local i_pat = t_i_pat
    local ret = 1
    for i = 1, len do
      div = div // (divisorparts[i].cnt + 1)
      local mul = i_pat // div
      i_pat = i_pat % div
      for j = 1, mul do
        ret = ret * divisorparts[i].p
      end
    end
    table.insert(t, ret)
  end
  table.sort(t)
  return t
end

local n, s = io.read("*n", "*n")
local lim = 316228

local function solve(x)
  local tn = n
  local c = 0
  while 0 < tn do
    c = c + tn % x
    tn = tn // x
  end
  return c == s
end

local found = false
for x = 2, lim - 1 do
  if solve(x) then
    found = true
    print(x)
    break
  end
end
if not found then
  if n == s then
    print(n + 1)
    found = true
  elseif s < n then
    local mul = n - s
    local primes = getprimes(lim)
    local divp = getdivisorparts(mul, primes)
    local divs = getdivisor(divp)
    for i = 1, #divs do
      local cand = divs[i]
      local a1 = mul // cand
      if a1 <= cand then
        local a0 = s - a1
        if 0 <= a0 and a0 <= cand then
          found = true
          print(cand + 1)
          break
        end
      end
    end
  end
  if not found then
    print(-1)
  end
end

Score : 500 points

Problem Statement

For integers b (b \geq 2) and n (n \geq 1), let the function f(b,n) be defined as follows:

f(b,n) = n, when n < b

f(b,n) = f(b,\,{\rm floor}(n / b)) + (n \ {\rm mod} \ b), when n \geq b

Here, {\rm floor}(n / b) denotes the largest integer not exceeding n / b,
and n \ {\rm mod} \ b denotes the remainder of n divided by b.

Less formally, f(b,n) is equal to the sum of the digits of n written in base b.
For example, the following hold:

f(10,\,87654)=8+7+6+5+4=30

f(100,\,87654)=8+76+54=138

You are given integers n and s.
Determine if there exists an integer b (b \geq 2) such that f(b,n)=s.
If the answer is positive, also find the smallest such b.

Constraints

1 \leq n \leq 10^{11}

1 \leq s \leq 10^{11}

n,\,s are integers.

Input

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

n
s

Output

If there exists an integer b (b \geq 2) such that f(b,n)=s, print the smallest such b.
If such b does not exist, print -1 instead.

Sample Input 1

87654
30

Sample Output 1

10

Sample Input 2

87654
138

Sample Output 2

100

Sample Input 3

87654
45678

Sample Output 3

-1

Sample Input 4

31415926535
1

Sample Output 4

31415926535

Sample Input 5

1
31415926535

Sample Output 5

-1


local mfl = math.floor
local mce, msq = math.ceil, math.sqrt
local function getprimes(x)
  local primes = {}
  local allnums = {}
  for i = 1, x do allnums[i] = true end
  for i = 2, x do
    if allnums[i] then
      table.insert(primes, i)
      local lim = mfl(x / i)
      for j = 2, lim do
        allnums[j * i] = false
      end
    end
  end
  return primes
end

local function getdivisorparts(x, primes)
  local prime_num = #primes
  local tmp = {}
  local lim = mce(msq(x))
  local primepos = 1
  local dv = primes[primepos]
  while primepos <= prime_num and dv <= lim do
    if x % dv == 0 then
      local asdf = {}
      asdf.p = dv
      asdf.cnt = 1
      x = x / dv
      while x % dv == 0 do
        x = x / dv
        asdf.cnt = asdf.cnt + 1
      end
      table.insert(tmp, asdf)
      lim = mce(msq(x))
    end
    if primepos == prime_num then break end
    primepos = primepos + 1
    dv = primes[primepos]
  end
  if x ~= 1 then
    local asdf = {}
    asdf.p, asdf.cnt = x, 1
    table.insert(tmp, asdf)
  end
  return tmp
end

local function getdivisor(divisorparts)
  local t = {}
  local pat = 1
  local len = #divisorparts
  local allpat = 1
  for i = 1, len do
    allpat = allpat * (1 + divisorparts[i].cnt)
  end
  for t_i_pat = 0, allpat - 1 do
    local div = allpat
    local i_pat = t_i_pat
    local ret = 1
    for i = 1, len do
      div = mfl(div / (divisorparts[i].cnt + 1))
      local mul = mfl(i_pat / div)
      i_pat = i_pat % div
      for j = 1, mul do
        ret = ret * divisorparts[i].p
      end
    end
    table.insert(t, ret)
  end
  table.sort(t)
  return t
end

local n, s = io.read("*n", "*n")
local lim = 316228

local function solve(x)
  local tn = n
  local c = 0
  while 0 < tn do
    c = c + tn % x
    tn = mfl(tn / x)
  end
  return c == s
end

local found = false
for x = 2, lim - 1 do
  if solve(x) then
    found = true
    print(x)
    break
  end
end
if not found then
  if n == s then
    print(n + 1)
    found = true
  elseif s < n then
    local mul = n - s
    local primes = getprimes(lim)
    local divp = getdivisorparts(mul, primes)
    local divs = getdivisor(divp)
    for i = 1, #divs do
      local cand = divs[i]
      local a1 = mfl(mul / cand)
      if a1 <= cand then
        local a0 = s - a1
        if 0 <= a0 and a0 <= cand then
          found = true
          print(cand + 1)
          break
        end
      end
    end
  end
  if not found then
    print(-1)
  end
end


s=io.read()
t={}
for i=1,#s do
    table.insert(t,s:sub(i,i))
end
table.sort(t)
x=table.concat(t,"")
if string.match(x,"E+N+S+W+") or string.match(x,"E+W+")==x or string.match(x,"N+S+")==x then
    print(x)
    print("Yes")
else
    print("No")
end

Score : 200 points

Problem Statement

Snuke lives on an infinite two-dimensional plane. He is going on an N-day trip.
At the beginning of Day 1, he is at home. His plan is described in a string S of length N.
On Day i(1 ≦ i ≦ N), he will travel a positive distance in the following direction:

North if the i-th letter of S is N

West if the i-th letter of S is W

South if the i-th letter of S is S

East if the i-th letter of S is E

He has not decided each day's travel distance. Determine whether it is possible to set each day's travel distance so that he will be back at home at the end of Day N.

Constraints

1 ≦ | S | ≦ 1000

S consists of the letters N, W, S, E.

Input

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

S

Output

Print Yes if it is possible to set each day's travel distance so that he will be back at home at the end of Day N. Otherwise, print No.

Sample Input 1

SENW

Sample Output 1

Yes

If Snuke travels a distance of 1 on each day, he will be back at home at the end of day 4.

Sample Input 2

NSNNSNSN

Sample Output 2

Yes

Sample Input 3

NNEW

Sample Output 3

No

Sample Input 4

W

Sample Output 4

No


local n = io.read("*n")
local cnt, rem = 0, 0
for i = 1, n do
  local a = io.read("*n")
  if a == 0 then
    rem = 0
  else
    a = a + rem
    cnt = cnt + a // 2
    rem = a % 2
  end
end
print(cnt)

Score : 400 points

Problem Statement

Snuke has a large collection of cards. Each card has an integer between 1 and N, inclusive, written on it.
He has A_i cards with an integer i.

Two cards can form a pair if the absolute value of the difference of the integers written on them is at most 1.

Snuke wants to create the maximum number of pairs from his cards, on the condition that no card should be used in multiple pairs. Find the maximum number of pairs that he can create.

Constraints

1 ≦ N ≦ 10^5

0 ≦ A_i ≦ 10^9 (1 ≦ i ≦ N)

All input values are integers.

Input

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

N
A_1
:
A_N

Output

Print the maximum number of pairs that Snuke can create.

Sample Input 1

4
4
0
3
2

Sample Output 1

4

For example, Snuke can create the following four pairs: (1,1),(1,1),(3,4),(3,4).

Sample Input 2

8
2
0
1
6
0
8
2
1

Sample Output 2

9


local function axiscomp(ary)
  local z = {}
  for i = 1, #ary do
    z[i] = ary[i]
  end
  table.sort(z)
  local ac, acinv = {}, {}
  local cur = 0
  for i = 1, #z do
    if not acinv[z[i]] then
      cur = cur + 1
      ac[cur] = z[i]
      acinv[z[i]] = cur
    end
  end
  return ac, acinv
end

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

local tc, tcinv = axiscomp(t)

local ret = 0
for i = 1, n, 2 do
  local v = tcinv[t[i]]
  if v % 2 == 0 then ret = ret + 1 end
end
print(ret)

Score : 600 points

Problem Statement

Snuke got an integer sequence of length N from his mother, as a birthday present. The i-th (1 ≦ i ≦ N) element of the sequence is a_i. The elements are pairwise distinct.
He is sorting this sequence in increasing order.
With supernatural power, he can perform the following two operations on the sequence in any order:

Operation 1: choose 2 consecutive elements, then reverse the order of those elements.

Operation 2: choose 3 consecutive elements, then reverse the order of those elements.

Snuke likes Operation 2, but not Operation 1. Find the minimum number of Operation 1 that he has to perform in order to sort the sequence in increasing order.

Constraints

1 ≦ N ≦ 10^5

0 ≦ A_i ≦ 10^9

If i ≠ j, then A_i ≠ A_j.

All input values are integers.

Input

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

N
A_1
:
A_N

Output

Print the minimum number of times Operation 1 that Snuke has to perform.

Sample Input 1

4
2
4
3
1

Sample Output 1

1

The given sequence can be sorted as follows:

First, reverse the order of the last three elements. The sequence is now: 2,1,3,4.

Then, reverse the order of the first two elements. The sequence is now: 1,2,3,4.

In this sequence of operations, Operation 1 is performed once. It is not possible to sort the sequence with less number of Operation 1, thus the answer is 1.

Sample Input 2

5
10
8
5
3
2

Sample Output 2

0


local mod = 1000000007
local function badd(x, y) return (x + y) % mod end
local mfl = math.floor

local function bmul(x, y)
  local x1, y1 = mfl(x / 31623), mfl(y / 31623)
  local x0, y0 = x - x1 * 31623, y - y1 * 31623
  return (x1 * y1 * 14122 + (x1 * y0 + x0 * y1) * 31623 + x0 * y0) % 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 n = io.read("*n", "*l")
local _u = io.read()
local slen = #_u
local dp1, dp2 = {}, {}
dp1[1] = 1
for i = 1, n do
  local src = i % 2 == 1 and dp1 or dp2
  local dst = i % 2 == 1 and dp2 or dp1
  for j = 1, i + 1 do
    dst[j] = 0
  end
  dst[1] = src[1]
  dst[2] = badd(dst[2], src[1] + src[1])
  for j = 2, i do
    dst[j - 1] = badd(dst[j - 1], src[j])
    dst[j + 1] = badd(dst[j + 1], src[j] + src[j])
  end
end

local tbl = n % 2 == 0 and dp1 or dp2
local ret = tbl[1 + slen]
local denom = modinv(modpow(2, slen))
print(bmul(ret, denom))

Score : 800 points

Problem Statement

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the 0 key, the 1 key and the backspace key.

To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:

The 0 key: a letter 0 will be inserted to the right of the string.

The 1 key: a letter 1 will be inserted to the right of the string.

The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.

Sig has launched the editor, and pressed these keys N times in total. As a result, the editor displays a string s. Find the number of such ways to press the keys, modulo 10^9 + 7.

Constraints

1 ≦ N ≦ 5000

1 ≦ |s| ≦ N

s consists of the letters 0 and 1.

Partial Score

400 points will be awarded for passing the test set satisfying 1 ≦ N ≦ 300.

Input

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

N
s

Output

Print the number of the ways to press the keys N times in total such that the editor displays the string s in the end, modulo 10^9+7.

Sample Input 1

3
0

Sample Output 1

5

We will denote the backspace key by B. The following 5 ways to press the keys will cause the editor to display the string 0 in the end: 00B, 01B, 0B0, 1B0, BB0. In the last way, nothing will happen when the backspace key is pressed.

Sample Input 2

300
1100100

Sample Output 2

519054663

Sample Input 3

5000
01000001011101000100001101101111011001000110010101110010000

Sample Output 3

500886057


Score : 100 points

Problem Statement

There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

Constraints

1≦N≦100

Input

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

N

Output

Print the necessary number of candies in total.

Sample Input 1

3

Sample Output 1

6

The answer is 1+2+3=6.

Sample Input 2

10

Sample Output 2

55

The sum of the integers from 1 to 10 is 55.

Sample Input 3

1

Sample Output 3

1

Only one child. The answer is 1 in this case.


local n = io.stdin:read("*n")
print(n * (n+1) // 2)


Score : 200 points

Problem Statement

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the 0 key, the 1 key and the backspace key.

To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:

The 0 key: a letter 0 will be inserted to the right of the string.

The 1 key: a letter 1 will be inserted to the right of the string.

The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.

Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter 0 stands for the 0 key, the letter 1 stands for the 1 key and the letter B stands for the backspace key. What string is displayed in the editor now?

Constraints

1 ≦ |s| ≦ 10 (|s| denotes the length of s)

s consists of the letters 0, 1 and B.

The correct answer is not an empty string.

Input

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

s

Output

Print the string displayed in the editor in the end.

Sample Input 1

01B0

Sample Output 1

00

Each time the key is pressed, the string in the editor will change as follows: 0, 01, 0, 00.

Sample Input 2

0BB1

Sample Output 2

1

Each time the key is pressed, the string in the editor will change as follows: 0, (empty), (empty), 1.


local a, b = io.read("*n", "*n")
if a * b <= 0 then print("Zero")
elseif 0 < a then
  print("Positive")
else
  if (b - a) % 2 == 1 then
    print("Positive")
  else
    print("Negative")
  end
end

Problem Statement

You are given two integers a and b (a≤b). Determine if the product of the integers a, a+1, …, b is positive, negative or zero.

Constraints

a and b are integers.

-10^9≤a≤b≤10^9

Partial Score

In test cases worth 100 points, -10≤a≤b≤10.

Input

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

a b

Output

If the product is positive, print Positive. If it is negative, print Negative. If it is zero, print Zero.

Sample Input 1

1 3

Sample Output 1

Positive

1×2×3=6 is positive.

Sample Input 2

-3 -1

Sample Output 2

Negative

(-3)×(-2)×(-1)=-6 is negative.

Sample Input 3

-1 1

Sample Output 3

Zero

(-1)×0×1=0.


local n,m=io.read("n","n")
local box={}
local red={}
for i=1,n do
    box[i]=1
    red[i]=false
end
red[1]=true

for i=1,m do
    local x,y=io.read("n","n")
    if red[x] then
        red[y]=true
    end
    box[x]=box[x]-1
    box[y]=box[y]+1
    if box[x]==0 then
        red[x]=false
    end
end

local counter=0
for i=1,n do
    counter=counter+(red[i] and 1 or 0)
end
print(counter)

Problem Statement

We have N boxes, numbered 1 through N. At first, box 1 contains one red ball, and each of the other boxes contains one white ball.

Snuke will perform the following M operations, one by one. In the i-th operation, he randomly picks one ball from box x_i, then he puts it into box y_i.

Find the number of boxes that may contain the red ball after all operations are performed.

Constraints

2≤N≤10^5

1≤M≤10^5

1≤x_i,y_i≤N

x_i≠y_i

Just before the i-th operation is performed, box x_i contains at least 1 ball.

Input

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

N M
x_1 y_1
:
x_M y_M

Output

Print the number of boxes that may contain the red ball after all operations are performed.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

2

Just after the first operation, box 1 is empty, box 2 contains one red ball and one white ball, and box 3 contains one white ball.

Now, consider the second operation. If Snuke picks the red ball from box 2, the red ball will go into box 3. If he picks the white ball instead, the red ball will stay in box 2.
Thus, the number of boxes that may contain the red ball after all operations, is 2.

Sample Input 2

3 3
1 2
2 3
2 3

Sample Output 2

1

All balls will go into box 3.

Sample Input 3

4 4
1 2
2 3
4 1
3 4

Sample Output 3

3


local read = setmetatable({}, {__index = function(t, k) local a = {} for i=1,#k do table.insert(a, '*'..string.sub(k, i, i)) end local r = io.read local u = table.unpack or unpack return function() return r(u(a)) end end})
read.N = function(N) local t={} for i=1,N do t[i]=read.n() end return t end
--
local N, L = read.nn()
local A = read.N(N)

local max = 0
local max_i = 0
for i=1,N-1 do
    local cur = A[i] + A[i+1]
    if cur > max then
        max = cur
        max_i = i
    end
end
if max < L then
    print("Impossible")
    return
else
    print("Possible")
    for i=1,max_i-1 do
        print(i)
    end
    for i=N-1,max_i,-1 do
        print(i)
    end
end

Problem Statement

We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i.

At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly:

Choose a (connected) rope with a total length of at least L, then untie one of its knots.

Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.

Constraints

2≤N≤10^5

1≤L≤10^9

1≤a_i≤10^9

All input values are integers.

Input

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

N L
a_1 a_2 ... a_n

Output

If it is not possible to untie all of the N-1 knots, print Impossible.

If it is possible to untie all of the knots, print Possible, then print another N-1 lines that describe a possible order to untie the knots. The j-th of those N-1 lines should contain the index of the knot that is untied in the j-th operation. Here, the index of the knot connecting piece i and piece i+1 is i.

If there is more than one solution, output any.

Sample Input 1

3 50
30 20 10

Sample Output 1

Possible
2
1

If the knot 1 is untied first, the knot 2 will become impossible to untie.

Sample Input 2

2 21
10 10

Sample Output 2

Impossible

Sample Input 3

5 50
10 20 30 40 50

Sample Output 3

Possible
1
2
3
4

Another example of a possible solution is 3, 4, 1, 2.


local n, l = io.read("*n", "*n")
local a = {}
for i = 1, n do
  a[i] = io.read("*n")
end
local pos = 0
for i = 1, n - 1 do
  if l <= a[i] + a[i + 1] then
    pos = i
    break
  end
end
if pos == 0 then
  print("Impossible")
else
  print("Possible")
  for i = 1, pos - 1 do
    print(i)
  end
  for i = n - 1, pos + 1, -1 do
    print(i)
  end
  print(pos)
end


local t = {}
for i = 1, 3 do
  table.insert(t, io.read("*n", "*n", "*n"))
end
table.sort(t)
if t[1] == 5 and t[2] == 5 and t[3] == 7 then
  print("YES")
else
  print("NO")
end

Score : 100 points

Problem Statement

Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.

To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Constraints

1≦A,B,C≦10

Input

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

A B C

Output

If it is possible to construct a Haiku by using each of the phrases once, print YES (case-sensitive). Otherwise, print NO.

Sample Input 1

5 5 7

Sample Output 1

YES

Using three phrases of length 5, 5 and 7, it is possible to construct a Haiku.

Sample Input 2

7 7 5

Sample Output 2

NO


local n, k = io.read("*n", "*n")
local t = {}
for i = 1, k do
  local z = io.read("*n")
  t[z] = true
end
while true do
  local len = #(tostring(n))
  local valid = true
  local v = n
  for i = 1, len do
    local tmp = v % 10
    if t[tmp] then valid = false break end
    v = v // 10
  end
  if valid then print(n) break end
  n = n + 1
end

Score : 300 points

Problem Statement

Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K.

She is shopping, and now paying at the cashier.
Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change).

However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money.

Find the amount of money that she will hand to the cashier.

Constraints

1 ≦ N < 10000

1 ≦ K < 10

0 ≦ D_1 < D_2 < … < D_K≦9

\{D_1,D_2,...,D_K\} ≠ \{1,2,3,4,5,6,7,8,9\}

Input

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

N K
D_1 D_2 … D_K

Output

Print the amount of money that Iroha will hand to the cashier.

Sample Input 1

1000 8
1 3 4 5 6 7 8 9

Sample Output 1

2000

She dislikes all digits except 0 and 2.

The smallest integer equal to or greater than N=1000 whose decimal notation contains only 0 and 2, is 2000.

Sample Input 2

9999 1
0

Sample Output 2

9999


n, k = io.read("*n", "*n")
ds = {}
for i = 1, k do ds[io.read("*n")] = true end
digits = {}
while(0 < n) do
    a = n % 10
    table.insert(digits, a)
    n = math.floor(n / 10)
end
needrestart = true
while(needrestart) do
    needrestart = false
    len = #digits
    for i = 1, len + 1 do
        val = digits[i]
        if(val == nil) then break end
        isok = (ds[val % 10] == nil)
        while(not isok)do
            needrestart = true
            for j = 1, i - 1 do digits[j] = 0 end
            val = val + 1
            isok = (ds[val % 10] == nil)
        end
        if(10 <= val) then
            needrestart = true
            for j = 1, i - 1 do digits[j] = 0 end
            val = val - 10
            if(i ~= len) then digits[i + 1] = digits[i + 1] + 1 else digits[i + 1] = 1 end
        end
        digits[i] = val
        if(needrestart) then break end
    end
end
len = #digits
for i = len , 1, -1 do
    io.write(digits[i])
end
io.write("\n")


local ior = io.input()
local n = ior:read("*n")
local t = {}
for i = 1, 2 * n do t[i] = ior:read("*n") end
table.sort(t)
local sum = 0
for i = 1, 2 * n, 2 do
  sum = sum + t[i]
end
print(sum)

Score : 200 points

Problem Statement

Snuke is having a barbeque party.

At the party, he will make N servings of Skewer Meal.

Example of a serving of Skewer Meal

He has a stock of 2N skewers, all of which will be used in Skewer Meal. The length of the i-th skewer is L_i.
Also, he has an infinite supply of ingredients.

To make a serving of Skewer Meal, he picks 2 skewers and threads ingredients onto those skewers.
Let the length of the shorter skewer be x, then the serving can hold the maximum of x ingredients.

What is the maximum total number of ingredients that his N servings of Skewer Meal can hold, if he uses the skewers optimally?

Constraints

1≦N≦100

1≦L_i≦100

For each i, L_i is an integer.

Input

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

N
L_1 L_2 ... L_{2N}

Output

Print the maximum total number of ingredients that Snuke's N servings of Skewer Meal can hold.

Sample Input 1

2
1 3 1 2

Sample Output 1

3

If he makes a serving using the first and third skewers, and another using the second and fourth skewers, each serving will hold 1 and 2 ingredients, for the total of 3.

Sample Input 2

5
100 1 2 3 14 15 58 58 58 29

Sample Output 2

135


local ior = io.input()
local n, x = ior:read("*n", "*n")
local sum = n
local a, b = x, n - x
if(a < b) then a, b = b, a end
while(true) do
  if(a % b == 0) then
    sum = sum + b * (2 * (a // b) - 1)
    break
  else
    local rem = a % b
    sum = sum + 2 * (a - rem)
    a, b = b, rem
  end
end
print(sum)

Score : 500 points

Problem Statement

Snuke is conducting an optical experiment using mirrors and his new invention, the rifle of Mysterious Light.

Three mirrors of length N are set so that they form an equilateral triangle.
Let the vertices of the triangle be a, b and c.

Inside the triangle, the rifle is placed at the point p on segment ab such that ap = X.
(The size of the rifle is negligible.)
Now, the rifle is about to fire a ray of Mysterious Light in the direction of bc.

The ray of Mysterious Light will travel in a straight line, and will be reflected by mirrors, in the same ways as "ordinary" light.
There is one major difference, though: it will be also reflected by its own trajectory as if it is a mirror!
When the ray comes back to the rifle, the ray will be absorbed.

The following image shows the ray's trajectory where N = 5 and X = 2.

It can be shown that the ray eventually comes back to the rifle and is absorbed, regardless of the values of N and X.
Find the total length of the ray's trajectory.

Constraints

2≦N≦10^{12}

1≦X≦N-1

N and X are integers.

Partial Points

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

Another 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 X

Output

Print the total length of the ray's trajectory.

Sample Input 1

5 2

Sample Output 1

12

Refer to the image in the Problem Statement section.
The total length of the trajectory is 2+3+2+2+1+1+1 = 12.