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 Python (3.8.2) (judge verdict: Accepted)

from math import *

if __name__ == '__main__':
    N = int(input())
    s = input().rstrip().split(' ')
    A = [int(elem) for elem in s]

    left = []
    total = sum(A)
    for elem in A:
        left.append(total - elem)
        total -= elem

    active = 1
    route = [min(active-A[0], left[0])]
    if route[0] <= 0: 
        if 0 == N and route[0] == 0:
            pass
        else:
            print(-1)
            exit()

    for i in range(1,N+1):
        active = active*2-A[i]
        route.append(min(active, left[i]))
        #print(route[i])
        if route[i] <= 0: 
            if i == N and route[i] == 0:
                break
            else:
                print(-1)
                exit()

    print(sum(route) + sum(A))