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)

INF = 100000
n = gets.to_i
nums = gets.split.map(&:to_i)
pow_memo = [1]
(0..INF).each do |i|
  if pow_memo[i] < 10 ** 500
    pow_memo[i + 1] = pow_memo[i] * 2
  else
    pow_memo[i + 1] = pow_memo[i]
  end
end

leaves = [1]
rests = [nums.sum]
0.upto(n - 1) do |from|
  tmp = rests[from]
  pow = pow_memo[from + 1]
  tmp = pow if tmp > pow
  tmp2 = (leaves[from] - nums[from]) * 2
  tmp = tmp2 if tmp > tmp2
  leaves[from + 1] = tmp
  if tmp < 1
    puts -1
    exit
  end
  rest = rests[from]
  rest -= nums[from + 1]
  if rest < 1 && from != n - 1
    puts -1
    exit
  end
  rests[from + 1] = rest
end

puts leaves.sum