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>
#define rep(i,n) for (int i = 0; i < n; ++i)
using namespace std;
using ll = long long;
const ll mod = 1000000007;//998244353;
int main() {
    cin.tie(nullptr); ios::sync_with_stdio(false);
    int n;cin >> n;
    vector<ll> a(n+1), b, c(n+2,0);
    rep(i,n+1)cin >> a[i];
    rep(i,n+1)c[n-i] += c[n-i+1] + a[n-i];
    b.push_back(1-a[0]);
    rep(i,n)
        b.push_back(min(2*b[i]-a[i+1], c[i+2]));
    ll ans = 0;
    rep(i,n+1){
        if (b[i]<0){
            cout << -1 << endl;
            return 0;
        }
        ans += a[i]+b[i];
    }
    cout << ans << endl;
}