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: Runtime Error)

if __name__ == '__main__':
    N = int(input().strip())
    A = list(map(int, input().strip().split(" ")))



    if A[0] != 1 and len(A) == 1:
        print(-1)

    tree = [0] * (sum([2 ** i for i in range(N + 1)]))
    for i, a in enumerate(A[::-1]):
        depth = len(A) - 1 - i
        depth_hierarchy_start_index = sum([2 ** i for i in range(depth)]) if depth != 0 else 0
        depth_hierarchy_end_index = sum([2 ** i for i in range(depth + 1)]) if depth != 0 else 1

        # keisho
        if depth != N:
            for i in range(depth_hierarchy_start_index, depth_hierarchy_end_index):
                if tree[2 * i + 1] != 0 or tree[2 * i + 2] != 0:
                    tree[i] = 2

        # 1 ume
        count = 0
        part = 2
        while True:
            for i in range(0, 2 ** depth, int(2 ** depth / part) if int(2 ** depth / part) >= 1 else 1):
                if tree[depth_hierarchy_start_index + i] == 0:
                    tree[depth_hierarchy_start_index + i] = 1
                    count += 1
                if count == a:
                    break
            part *= 2
            if count == a:
                break
            if int(2 ** depth / part) < 1 and count < a:
                print(-1)
                break


    if not int(2 ** depth / part) < 1 and not  count < a:
        print(sum([1 if i != 0 else 0 for i in tree]))