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;
int main(){
    int n;
    cin>>n;
    int a[n+1];
    for(int i=0;i<=n;i++)cin>>a[i];
    long long b[n+1],wa[n+1];
    b[0]=1-a[0];
    if(b[0]<0){
        cout<<"-1"<<endl;
        return 0;
    }
    wa[0]=a[0];
    for(int i=1;i<=n;i++)wa[i]=wa[i-1]+a[i];
    for(int i=1;i<=n;i++){
        b[i]=min(2*b[i-1]-a[i],wa[n]-wa[i]);
        if(b[i]+a[i]<b[i-1] || b[i]<0){
            cout<<"-1"<<endl;
            return 0;
        }
    }
    long long x=0;
    for(int i=0;i<=n;i++)x+=a[i]+b[i];
    cout<<x<<endl;
    return 0;
}
