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>

int main(){
    int N;
    std::cin >> N;
    std::vector< int > A(N+1);
    long long total = 0;
    for(int i=0; i<N+1; i++){
        std::cin >> A[i];
        total += A[i];
    }
    
    long long cnt_d = 1;
    long long cnt = 0;
    for(int i=0; i<N+1; i++){
        if(cnt_d < A[i]){
            std::cout << -1 << std::endl;
            return 0;
        }
        cnt += cnt_d;
        cnt_d = (cnt_d - (long long)A[i]) * 2;
        total -= A[i];
        if(cnt_d > total){
            cnt_d = total;
        }
    }
    
    std::cout << cnt << std::endl;
    
    return 0;
}

