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;

const int N = 100005;

int n;
long long a[N], l[N], r[N], ans = 0;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    for (int i = 0; i <= n; i++) {
        cin >> a[i];
    }
    l[n] = r[n] = a[n];
    for (int i = n - 1; i >= 0; i--) {
        l[i] = (l[i + 1] + 1) / 2;
        r[i] = r[i + 1];
        l[i] += a[i]; r[i] += a[i];
    }
    if (l[0] > 1) {
        return cout << -1, 0;
    }
    long long cur = 1;
    for (int i = 0; i <= n; i++) {
        ans += cur;
        cur = min(r[i + 1], (cur - a[i]) * 2);
    }
    cout << ans;
}
