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;

int main(){
    ll N;
    cin>>N;
    vector<ll> A(N+10);
    ll sum = 0;
    for(int i=0; i<=N; i++){
        cin>>A[i];
        sum += A[i];
    }

    ll cur = 1;
    ll ans = 1;

    for(int i=0; i<=N; i++){
        cur -= A[i];
        if(cur < 0){
            ans = -1;
            break;
        }

        sum -= A[i];
        cur = min(sum,cur*2);
        ans += cur;
    }

    cout<<ans<<endl;

    return 0;
}