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;

typedef long long ll;

int n;
int a[100005];
ll mx[100005];

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n;
    for(int i=0;i<=n;i++){
        cin>>a[i];
    }
    for(int i=n;i>=0;i--){
        mx[i]=mx[i+1]+a[i];
    }
    ll tot=0,cur=1;
    for(int i=0;i<=n;i++){
        tot+=cur;
        cur-=a[i];
        if(cur<0||cur>mx[i+1]){
            cout<<-1<<endl;
            return 0;
        }
        cur=min(cur<<1,mx[i+1]);
    }
    cout<<tot<<endl;

    return 0;
}