Problem:
Score : 600 points

Problem Statement

We have one slime.

You can set the health of this slime to any integer value of your choice.

A slime reproduces every second by spawning another slime that has strictly less health. You can freely choose the health of each new slime. The first reproduction of our slime will happen in one second.

Determine if it is possible to set the healths of our first slime and the subsequent slimes spawn so that the multiset of the healths of the 2^N slimes that will exist in N seconds equals a multiset S.

Here S is a multiset containing 2^N (possibly duplicated) integers: S_1,~S_2,~...,~S_{2^N}.

Constraints

All values in input are integers.

1 \leq N \leq 18

1 \leq S_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
S_1 S_2 ... S_{2^N}

Output

If it is possible to set the healths of the first slime and the subsequent slimes spawn so that the multiset of the healths of the 2^N slimes that will exist in N seconds equals S, print Yes; otherwise, print No.

Sample Input 1

2
4 2 3 1

Sample Output 1

Yes

We will show one way to make the multiset of the healths of the slimes that will exist in 2 seconds equal to S.

First, set the health of the first slime to 4.

By letting the first slime spawn a slime whose health is 3, the healths of the slimes that exist in 1 second can be 4,~3.

Then, by letting the first slime spawn a slime whose health is 2, and letting the second slime spawn a slime whose health is 1, the healths of the slimes that exist in 2 seconds can be 4,~3,~2,~1, which is equal to S as multisets.

Sample Input 2

2
1 2 3 1

Sample Output 2

Yes

S may contain multiple instances of the same integer.

Sample Input 3

1
1 1

Sample Output 3

No

Sample Input 4

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

Sample Output 4

No

Solution in LuaJIT (2.0.4) (judge verdict: Wrong Answer)

local mfl, mce = math.floor, math.ceil
local mmi, mma = math.min, math.max
local bls, brs = bit.lshift, bit.rshift
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.lower_bound = function(self, val)
  local ret, retpos = self.emptyvalue, 0
  local stagenum = self.stagenum
  local stage, l, r = 1, 1, bls(1, stagenum - 1)
  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 l + sz <= r then stage, l = stage + 1, l + sz
      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.setValue = function(self, idx, value)
  self.stage[self.stagenum][idx] = value
  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
SegTree.new = function(n, func, emptyvalue)
  local obj = {}
  setmetatable(obj, {__index = SegTree})
  obj:create(n, func, emptyvalue)
  return obj
end
local inf = 1000000007
local n = io.read("*n")
local tot = bls(1, n)
local t = {}
local idx = {}
local childcnt = {}
local capacity = {}
for i = 1, tot do
  t[i] = io.read("*n")
  idx[i] = i
  childcnt[i] = 0
  capacity[i] = -1
end
table.sort(t)
local parent = {0}
local len = {0}
for i = 1, n do
  local sn = bls(1, i - 1)
  for j = 1, sn do
    childcnt[j] = childcnt[j] + 1
    parent[j + sn] = j
    len[j + sn] = len[j] + 1
  end
end
table.sort(idx, function(a, b) return len[a] > len[b] end)
-- print(table.concat(parent, " "))
-- print(table.concat(len, " "))
-- print(table.concat(idx, " "))

local st = SegTree.new(tot, mma, -inf)
for i = 1, tot do
  if childcnt[i] == 0 then
    st:setValue(i, capacity[i])
  end
end
local valid = true
for i = 1, tot do
  local pos = st:lower_bound(-t[i])
  if tot < pos then valid = false break end
  st:setValue(pos, -inf)
  if i ~= tot then
    local p = parent[pos]
    capacity[p] = mmi(capacity[p], -t[i] - 1)
    childcnt[p] = childcnt[p] - 1
    if childcnt[p] == 0 then
      st:setValue(p, capacity[p])
    end
  end
end
print(valid and "Yes" or "No")