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 PyPy3 (7.3.0) (judge verdict: Accepted)

from math import log2

N = int(input())
A = [int(i) for i in input().split()]

def solve() :
  m, M = A[-1], A[-1]
  max1 = [0] * (N + 1)
  max1[-1] = M
  for i in range(N - 1, -1, -1) :
    m = (m + 1) // 2 + A[i]
    M = M + A[i]
    max1[i] = M
  if not m <= 1 <= M :
    return -1
  
  ne = 1 - A[0]
  max2 = [0] * (N + 1)
  max2[0] = 1
  for i in range(1, N + 1) :
    ne = ne * 2 - A[i]
    max2[i] = ne + A[i]
    
  return sum(min(max1[i], max2[i])for i in range(N + 1))
  
print(solve())