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 C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#define ll long long
ll min(ll a, ll b){return a < b ? a : b;}
int main(){
    int n;
    scanf("%d",&n);
    ll a[n + 1], b[n + 1];
    for(int i = 0; i <= n; i++)
        scanf("%lld",&a[i]);
    b[0] = 1;
    for(int i = 0; i <= n; i++){
        if(b[i] < a[i]){
            puts("-1");
            return 0;
        }
        if(i == n)
            break;
        b[i] -= a[i];
        b[i + 1] = min(1LL << 60 , b[i] * 2);
    }
    ll ans = a[n], now = a[n];
    for(int i = n - 1; i >= 0; i--){
        ans += min(b[i], now) + a[i];
        now += a[i];
    }
    printf("%lld\n",ans);
    return 0;
}