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 ll long long
using namespace std;
ll a[100009],R[100009];
int main(){
    ll n;
    cin >> n;
    for(ll i = 0; i <= n; i++){
        cin >> a[i];
    }
    ll r = 1;
    ll inf = 1e18;
    for(ll i = 0; i < n; i++){
        if(a[i] > r) {cout << -1 <<endl; return 0;}
        R[i] = r;
        r = (r - a[i]) * 2;
        if(r > inf) r = inf;
    }
    if(a[n] > r){
        cout<<-1<<endl;
        return 0;
    }
    ll x = a[n];
    ll ans = 0;
    for(ll i = n; i >= 0; i--){
        ans += x;
        x = min(R[i-1],  x+ a[i-1]);
    }
    cout<<ans<<endl;

}

