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>
#define ref(i,n) for(ll i=0;i<n;i++)
using namespace std;
typedef long long ll;
#define INF 11111111111111111

ll a=0,n,t=1,b[1000000],ans=0,c[1000000];
int main(){

    cin>>n;
    ref(i,n+1){
        cin>>a;
        b[i]=a;
        c[i]=t-a;
        if(t>=INF) continue;
        if(t>=a) t=(t-a)*2;
        else{cout<<-1; return 0;}
    }
    a=0;
    for(ll i=n;i>0;i--){
            ans+=b[i];
        if(c[i-1]>=b[i]) {b[i-1]+=b[i];}
        else {b[i-1]+=c[i-1];}
        //cout<<ans;
    }
    ans+=b[0];
    cout<<ans;
    return 0;

}

