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>
#include <vector>

int main() {
    int N;
    std::cin >> N;

    std::vector<long long> A(N + 1);
    for (int i = 0; i <= N; i++) std::cin >> A[i];

    std::vector<long long> sum(N + 1);
    sum[N] = A[N];
    for (int i = N - 1; i >= 0; i--) {
        sum[i] = sum[i + 1] + A[i];
    }

    long long ans = 0, res = 0;
    for (int i = 0; i <= N; i++) {
        res = std::min(sum[i], res * 2 - A[i] + (i == 0 ? 1 : 0));

        if (res < 0) {
            std::cout << "-1\n";
            return 0;
        }

        ans += std::min(sum[i], res + A[i]);
    }
    std::cout << ans << "\n";
}