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 <iostream>
using namespace std;

int n, a[100005];
long long b, tot, ans;

int main() {
    cin >> n;
    for (int i = 0; i <= n; i++){
        cin >> a[i];
        tot += a[i];
    }
    b = 1;
    bool flag = true;
    for (int i = 0; i <= n; i++){
        if (b < a[i]){
            flag = false;
            break;
        }
        ans += b;
        b -= a[i];
        tot -= a[i];
        b = min(b*2, tot);
    }
    if (!flag) cout << -1 << "\n";
    else cout << ans << "\n";
}
