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>
#define rep(i, n) for (int i = 0; i < n; i++)
typedef long long ll;
using namespace std;

int main() {
    int n;
    cin >> n;
    ll a[n + 1], max[n + 1], min[n + 1];
    rep(i, n + 1) cin >> a[i];
    max[n] = min[n] = a[n];

    for (int i = n - 1; i >= 0; i--) {
        max[i] = max[i + 1] + a[i];
        min[i] = (min[i + 1] + 1) / 2 + a[i];
    }

    if (min[0] > 1) {
        cout << -1 << endl;
        return 0;
    }

    ll ans = 0, p = 1;
    rep(i, n + 1) {
        ans += p;
        p = (p - a[i]) * 2;
        if (p > max[i + 1]) p = max[i + 1];
    }

    cout << ans << endl;
    return 0;
}
