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 N = 1e5+10;
const ll INF = 1e18;
int n,f;
ll a[N],l[N],r[N],ans;
int main(){
    scanf("%d",&n); n++;
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    ll o=1;
    for(int i=1;i<=n;++i){
        l[i]=o-a[i]; o=l[i]*2>INF?INF:l[i]*2;
        //printf("%lld %lld\n",o,l[i]);
        if(l[i]<0){
            f=1; break;
        }
    }
    if(f){
        printf("-1\n");
        return 0;
    }
    for(int i=n;i>=1;--i)
        l[i]=min(l[i],l[i+1]+a[i+1]);
    for(int i=1;i<=n;++i){
        ans+=l[i]+a[i];
    }
    printf("%lld\n",ans);
}