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 <iostream>
#include <string>
using namespace std;
typedef long long ll;

const ll INF = 100000000000000000;

int main()
{
    int n;
    cin >> n;
    ll a[100005];
    for(int i = 0; i <= n; i++) cin >> a[i];
    ll ans = 0;
    ll s = 0;
    ll b[100005];
    b[0] = 1;
    for(int i = 0; i <= n; i++){
        b[i] -= a[i];
        if(b[i] < 0){
            cout << -1 << endl;
            return 0;
        }
        b[i + 1] = min(INF, b[i] * 2);
    }
    for(int i = n; i >= 0; i--){
        s = min(b[i], s) + a[i];
        ans += s;
    }
    cout << ans << endl;
}
