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;
typedef long long ll;


int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    ll totalLeaves = 0;
    vector<ll> a(n+1);
    for (auto &x: a) {
        cin >> x;
        totalLeaves += x;
    }

    ll cnt = 1;
    ll ans = 0;
    for (int i = 0; i <= n; ++i) {
        if (cnt < a[i]) {
            ans = -1;
            break;
        }
        ans += cnt;
        totalLeaves -= a[i];
        cnt -= a[i];
        cnt = min(cnt * 2, totalLeaves);
    }

    cout << ans << '\n';
}