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.

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.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264
#include <bits/stdc++.h>
using namespace std;

int n;
long long arr[100001];
long long ssum[100002];

int main(void) {
    scanf("%d",&n);
    for(int i=0;i<=n;i++) {
        scanf("%lld",&arr[i]);
    }
    ssum[n]=arr[n];
    for(int i=n-1;i>=0;i--) {
        ssum[i]=ssum[i+1]+arr[i];
    }
    long long tree=1;
    long long ret=0;
    for(int i=0;i<=n;i++) {
        ret+=tree;
        tree-=arr[i];
        if (tree<0) {
            printf("-1");
            return 0;
        }
        tree=min(tree*2,ssum[i+1]);
    }
    printf("%lld",ret);
}