Problem:
Score : 600 points

Problem Statement

Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

Constraints

0 \leq N \leq 10^5

0 \leq A_i \leq 10^{8} (0 \leq i \leq N)

A_N \geq 1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 A_2 \cdots A_N

Output

Print the answer as an integer.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

def calcpow(a, n)
  return 1 if n == 0
  exp = Math.log(n, 2).floor
  rep = calcpow_cl(a, n, exp)
  return rep[0]
end

def calcpow_cl(a, n, exp)
  if exp < 0
    rep = [1, a]
  else
    if n >= 2**exp
      rep = calcpow_cl(a, n-2**exp, exp-1)
      rep[0] = rep[0]*rep[1]
    else
      rep = calcpow_cl(a, n, exp-1)
    end
    rep[1] = rep[1]**2
  end
  return rep
end


n = gets.to_i
a = gets.split.map(&:to_i)

n > 0 && a[0] == 1 ? flag = 0 : flag = 1
for i in 1..n
  i-1 <= 30 ? all = calcpow(2,i-1) : all = 10**8
  if (all - a[i-1])*2 < a[i]
    flag = 0
    break
  end
end 
if flag == 0
  puts -1
  exit
end

able_max = Array.new(n+1)
able_min = Array.new(n+1)
able_max[0] = 1
able_min[0] = 1

for d in 1..n
  able_max[d] = (able_max[d-1]-a[d-1])*2
end

cnt = Array.new(n+1,0)
cnt[n] = [able_max[n], a[n]].min
for i in 1..n
  d = n-i
  max = [cnt[d+1]+a[d], able_max[d]].min
  min = [(cnt[d+1]+1)/2+a[d], able_max[d]].min
  if max >= min
    cnt[d] = max
  else
    flag = 0
    break
  end
end

ans = 0
cnt.each{|e| ans += e}
if flag == 1
  puts ans
else
  puts -1
end