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;
#define finish(x) return cout << x << endl, 0
#define ll long long

int n;
vector <int> a;

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    a.resize(n + 1);
    for(auto &i : a) cin >> i;
    ll cur = 1;
    ll ans = 0;
    ll tot = 0;
    for(auto &i : a) tot += i;
    for(int i = 0 ; i <= n ; i++){
        ans += cur;
        tot -= a[i];
        cur -= a[i];
        if(cur < 0) finish(-1);
        cur *= 2;
        cur = min(cur, tot);
    }
    cout << ans << endl;
}
