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;

int n;

void solve(){
    int i,j;
    string str;
    cin >> n;
    vector<long long> a(n+1),sa(n+2);
    for(i = 0;i <= n;i++) cin >> a[i];
    sa[n+1] = 0;
    for(i = n;i >= 0;i--) sa[i] = a[i] + sa[i+1];
    long long ans = 0;
    long long node = 1;
    for(i = 0;i < n;i++){
        if(a[i] >= node) {
            cout << -1 << endl;
            return;
        }
        ans += node;
        node = 2*(node-a[i]);
        if(node > sa[i+1]) node = sa[i+1];
    }
    if(a[i] == node) cout << ans+node << endl;
    else cout << -1 << endl;
}

int main(){
    solve();
}
