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 static ll INF = 1e15;
// __uint128_t

int main(){
    ll N; cin >> N;
    ll SUM = 0;
    vector<ll> C(N+1); for(int i = 0; i < N+1; i++) cin >> C[i];
    if(C[N] == 0 ) {cout << -1 << endl; return 0;}
    vector<ll> P(N+1); P[N] = 0;
    
    for(int i = N-1; i >= 0; i--){
        P[i] = C[i+1] + P[i+1];
    }


    ll temp = 1;
    for(int i = 0; i <= N; i++){
        P[i] = min(temp - C[i], P[i]); temp = 2*P[i];
        if(temp <= -1) {cout << -1 << endl;return 0;}

    }
    for(int i = 0; i < N+1; i++) SUM += C[i]+P[i];
    cout << SUM << endl;
}
