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: Accepted)

lines = []
while line = gets
    lines << line.chomp.split(' ').map(&:to_i)
end

N = lines[0][0]
leaves = []
for i in 0..N do
  leaves.push(lines[1][i])
end

max_bi = -1  # max bification
for i in 0..N do
  max_bi += leaves[i]
end

def max_nodes(leaves, max_bi)
  remain_bi = max_bi
  nodes = [1]       #always one node with depth 0
  for i in 0..N-1 do
    if nodes[i] < leaves[i] then
      return [-1]
    end
    num_fin = leaves[i]
    if remain_bi >= nodes[i] - num_fin then
      num_bi = nodes[i] - num_fin
      remain_bi -= nodes[i] - num_fin
      num_sin = 0
    else
      num_bi = remain_bi
      remain_bi = 0
      num_sin = nodes[i] - num_fin - num_bi
    end
    next_nodes = num_bi * 2 + num_sin
    nodes.push(next_nodes)
  end
  if nodes[N] < leaves[N] then
    return [-1]
  else
    return nodes
  end
end
  
num_nodes = max_nodes(leaves, max_bi)
ans = 0
for i in 0..(num_nodes.length-1) do
  ans += num_nodes[i]
end
print ans