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;
const int max_n=100005;
int n,a[max_n];
ll suf[max_n];
ll ans[max_n];
int main(){
    scanf("%d",&n);
    for(int i=0;i<=n;i++)scanf("%d",a+i);
    for(int i=n;i>=0;i--)suf[i]=suf[i+1]+a[i];
    ans[0]=1;
    for(int i=1;i<=n;i++){
        //ans[i]=min(ans[i-1]*2)
        ans[i]=min((ans[i-1]-a[i-1])*2,suf[i]);
    }
    for(int i=0;i<=n;i++)if(ans[i]<a[i])return printf("-1\n"),0;
    ll res=0;
    for(int i=0;i<=n;i++)res+=ans[i];
    printf("%lld\n",res);
    return 0;
}