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
/*
 * Author       : YangDavid
 * Created Time : 2020.05.30 20:21:54
 */

#include<bits/stdc++.h>
#define rep(i, n) for(int i = 1; i <= n; ++i)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
int main() {
    int n; scanf("%d", &n);
    vector<int> a(n + 1);
    ll tot = 0, fr = 1, ans = 0;
    for(auto &g : a)
        scanf("%d", &g), tot += g;
    for(auto ct : a) {
        if(fr < ct) { puts("-1"); return 0; }
        ans += ct, tot -= ct;
        ll cho = min(fr-ct, tot);
        ans += cho;
        fr = cho * 2;
    }
    printf("%lld\n", ans);


    return 0;
}

