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;
typedef long long ll;
int n;
ll a[100005];
ll p[100005];
ll S[100005];
ll ans=  0;
int main(){
    scanf("%d",&n);
    for (int i = 0; i <= n; i++){
        scanf("%lld",&a[i]);
    }
    for (int i = n; i >= 0; i--){
        p[i] = p[i+1]+a[i];
    }
    for (int i = 0; i <= n; i++){
        S[i] = (i==0?1:2*S[i-1])-a[i];
        S[i] = min(S[i],p[i+1]);
        ans += S[i]+a[i];
        if (S[i] < 0){
            printf("-1\n");
            return 0;
        }
        //printf("%d %lld\n",i,S[i]);
    }
    printf("%lld",ans);

}
