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: Wrong Answer)

N = int(input())
A = list(map(int,list(input().split())))

import math
count = 0
prev = 0
for n in range(N+1):
    depth = N - n
    print("depth:",depth)
    now = A.pop(depth)
    print("now:",now)
    border = 2**depth - sum([A[depth-i-1]*2**(i+1) for i in range(len(A))])
    print("border:",border)
    if now+math.ceil(prev/2) > border:
        count = -1
        break

    print("now:",now)
    if now+prev <= border:
        now += prev
    else:
        now = border
    print("now:",now)
    print("prev:",prev)
    count += now
    prev = now
    print("prev:",prev)
    print("count:",count)
    print("\n")