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(){
  long long n, a[100009], sm[100009];
  cin >> n;
  for(int i=0;i<=n;i++)cin >> a[i];
  sm[n] = a[n];
  for(int i=n-1;i>=0;i--)sm[i] = sm[i+1] + a[i];
  long long cnt = 0, mx=1;
  for(int i=0;i<=n;i++){
    if(mx<a[i]){
      cout << -1 << endl;
      return 0;
    }
    cnt += a[i];
    if(sm[i]<=mx){
      mx = sm[i]-a[i];
    }
    else{
      mx -= a[i];
    }
    cnt += mx;
    mx *= 2;
  }
  cout << cnt << endl;
}

