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() {
    int n; cin >> n;
    vector<ll> a(n+1);
    ll sum=0;
    cin >> a[0];
    for (int i=1; i<=n; ++i) {
        cin >> a[i];
        sum+=a[i];
    }
    if (a[n]==0) {
        cout << -1 << endl;
        return 0;
    }
    ll ans=0, cur=1;
    for (int i=0; i<=n; ++i) {
        sum-=a[i];
        ans+=cur;
        if (cur<a[i]) {
            cout << -1 << endl;
            return 0;
        }
        cur=min(cur*2-a[i]*2, sum);
    }
    cout << ans;
    return 0;
}