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;

int main(){
    long N;
    cin >> N;
    long A[N + 1];
    for(long i = 0; i < N + 1; i++)
        cin >> A[i];
    long B[N + 1];
    B[N] = A[N];
    for(long i = N - 1; i >= 0; i--){
        B[i] = B[i + 1] + A[i];
    }
    long ans = 0, a = 1;
    bool sw = true;
    for(long i = 0; i < N; i++){
        if(a < A[i]){
            sw = false;
            break;
        }
        ans += a;
        a = min(2 * (a - A[i]), B[i + 1]);
    }
    if(a != A[N]) sw = false;
    ans += A[N];
    if(sw) cout << ans << endl;
    else cout << -1 << endl;
    
    return 0;
}