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<cstdio>
#include<algorithm>
using namespace std;
int n, w[101000];
long long C[101000], INF = 1e15, S[101000], res;
int main(){
    int i;
    scanf("%d",&n);
    for(i=0;i<=n;i++){
        scanf("%d",&w[i]);
    }
    C[0]=1-w[0];
    for(i=1;i<=n;i++){
        C[i]=min(C[i-1]*2-w[i],INF);
    }
    for(i=0;i<=n;i++){
        if(C[i]<0){
            puts("-1");
            return 0;
        }
    }
    S[n]=w[n];
    for(i=n-1;i>=0;i--){
        S[i] = min(C[i],S[i+1])+w[i];
    }
    for(i=0;i<=n;i++){
        res+=S[i];
    }
    printf("%lld\n",res);
}
