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;
using i64 = long long;
#define endl "\n"

int main()
{
  i64 N;
  cin >> N;
  vector<i64> A(N + 1);
  for (i64 i = 0; i <= N; i++)
    cin >> A[i];
  vector<i64> sum = A;
  for (i64 i = N; 0 < i; i--)
    sum[i - 1] += sum[i];
  i64 ans = 0, node = 1;
  for (i64 i = 0; i < N; i++)
  {
    ans += node;
    if (node - A[i] < 0)
    {
      cout << -1 << endl;
      return 0;
    }
    node = min((node - A[i]) * 2, sum[i + 1]);
  }
  ans += node;
  if (node - A[N] == 0)
    cout << ans << endl;
  else
    cout << -1 << endl;
  return 0;
}
