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)

n = gets.to_i
a = gets.split.map(&:to_i).reverse
d = a.length
limit_by_bottom = [a.first, a.first]
q = []
a.each.with_index(1) do |ai, idx|
  di = d - idx
  # 下の深さからの制約（この階層のleaf数にプラスして下の階層のノード数が必要）
  bl = idx == 1 ? limit_by_bottom : [(limit_by_bottom[0]+ai).divmod(2).sum, limit_by_bottom[1]+ai]
  # この深さに対する制約
  cl = [idx == 1 ? ai : ai + 1, 2**di] # 存在可能なノード数の制約
  # 新制約
  nl = [[bl[0], cl[0]].max, [bl[1], cl[1]].min]

  # 制約が矛盾したら終了
  if nl[0] > nl[1]
    puts '-1'
    exit 0
  end

  # 矛盾しなければ上の深さに対する制約を設定して次へ
  q << [nl, ai]
  limit_by_bottom = nl
end
count = 0
lt = nil
q.reverse.each do |qi|
  if lt
    n = [lt[1], qi[0][1]].min
    count += n
    lt = [n - qi[1], (n - qi[1]) * 2]
  else
    count += qi[0].max
    lt = [qi[0].max - qi[1], (qi[0].max - qi[1]) * 2]
  end
end
puts count