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)

import sys 
sys.setrecursionlimit(2*100000) 

def dfs(i,pre,n,a,ans):
    if (pre*2 < a[i]): return -1
    ans[i] += a[i]
    if (i >= n):
        return a[i]

    pre = pre*2 - a[i]
    u = dfs(i+1,pre,n,a,ans);
    if (u<0 or pre*2 < u): return -1;
    ans[i] += min(pre, u)
    return min(pre, u) + a[i]



n = int(input())
a = list(map(int,input().split()))

if (n == 0):
    if (a[0] == 1):
        print(1)
    else:
        print(-1)
    exit()

ans = [1] + [0]*n
u = dfs(1,1,n,a,ans)
if (a[0] != 0): u = -1

if (u >= 0):
    print(sum(ans))
else:
    print(-1)