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() {
    using namespace std;
    using ll = long long;

    int N; std::cin >> N;
    std::vector<ll> A(N + 1), B(N + 1);
    for (int i = 0; i <= N; ++i) cin >> A[i];
    B[N] = A[N];
    for (int i = N - 1; i >= 0; --i) B[i] = B[i + 1] + A[i];
    B[0] = 1;
    for (int i = 1; i <= N; ++i) B[i] = min(B[i], 2 * (B[i - 1] - A[i - 1]));
    ll ans = 0;
    for (int i = 0; i <= N; ++i) {
        if (B[i] < A[i]) {
            cout << -1 << endl;
            return 0;
        }
        ans += B[i];
    }
    cout << ans << endl;
    return 0;
}
