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 <iostream>

using namespace std;

int main()
{
  int n;
  int *a;
  long int total_a = 0;
  long int e = 1;
  long int ans = 0;
  
  cin >> n;
  a = (int*)malloc(( n + 1 ) * sizeof( int ));
  if ( NULL == a ) { return -1; }
  for ( int i = 0 ; i <= n ; ++i ) { cin >> a[i]; total_a += a[i]; }
  
  for ( int i = 0 ; i <= n ; ++ i ) {
    ans += e;
    if ( e < a[i] ) { cout << -1 << endl; return 0; }
    e = 2 * ( e - a[i] );
    total_a -= a[i];
    e = min( e, total_a );
  }
  
  cout << ans << endl;
}