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;
  long int nord, allsum, temp;
  cin >> n;
  int a[n+1];
  allsum = 0;
  for(int i=0; i<=n; i++){
    cin >> a[i];
    allsum = allsum + a[i];
  }
  temp = 1;
  nord = allsum;
  for(int i=0; i<=n; i++){
    if(a[0] >= 2 ){
      nord = -1;
      break;
    }
    if(a[i] > temp*2){
      nord = -1;
      break;
    }
    else{
      allsum = allsum - a[i];
      if(i==0){
        temp = 1 - a[i];
        nord = nord + temp;
      }
      else{
        temp = min(temp*2 - a[i], allsum);
        nord = nord + temp;
      }
    }
  }
  cout << nord << endl;
}