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)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;

int main() {
    ll n;
    cin >> n;

    vector<ll> a(n+1);
    ll sum = 0;
    rep(i, n+1) { cin >> a[i]; sum += a[i]; }

    ll ans = 0;
    ll lim = 1;
    rep(i, n+1) {
        ans += lim;
        //printf("ans=%d, sum=%d, lim=%d\n", ans, sum, lim);
        sum -= a[i];
        lim = min((lim-a[i])*2, sum);
        if (lim < 0) {
            ans = -1; break;
        }
    }
    cout << ans << endl;
}
