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 i,j,k,l,m,n;

    cin >> n;

    ll ar[n+10],fir[n+10];

    for(i= 0 ; i <= n; i++)
        cin>>ar[i];

    ll ekn = 0,sum = 0;

    fir[n] = ar[n];

    for(i = n-1 ; i >= 0 ; i--)fir[i] = ar[i] + fir[i+1];

    for(i=0; i<=n; i++)
    {

       if(!i)ekn = 1;
       else
       {
           ekn = ekn*2;
       }
       ekn = min(ekn,fir[i]);
       sum += ekn;

       ekn -= ar[i];

       if(ekn < 0)return cout << -1 << endl,0;

    }

    cout << sum << endl;


}
