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 main()
{
    ll n;
    cin>>n;
    int i;
    ll A[n+1];
    for(i=0;i<=n;i++)
    cin>>A[i];
    ll B[n+1];
    ll ans=0;
    B[n]=A[n];
    for(i=n-1;i>=0;i--)
    B[i]=B[i+1]+A[i];
    B[0]=1;
    for(i=1;i<=n;i++)
    {
        B[i]=min(B[i],2*(B[i-1]-A[i-1]));
    }
    for(i=0;i<=n;i++)
    {
        if(B[i]<A[i])
        {
            ans=-1;
            break;
        }
        ans+=B[i];
    }
    cout<<ans<<"\n";
    return 0;
}