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;
long long n,c,i,a[100008],b[100008];
int main() {
    for(scanf("%lld",&n); i<=n; scanf("%lld",a+i++));
    for(c=1; i--; b[i]=a[i]+b[i+1],a[i]+=a[i+1]+1>>1);
    if(a[0]!=1)
        printf("-1\n"),exit(0);
    for(i=0; i++<n; c+=a[i]=min(a[i-1]<<1,b[i]),a[i]+=b[i+1]-b[i])
        if((a[i-1]+a[i-1])<a[i])
            printf("-1\n"),exit(0);
    printf("%lld\n",c),exit(0);
}