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 llint;

int main() {
  int n;
  cin >> n;
  vector<llint> a(n + 1);
  for (int i = 0; i <= n; i++) {
    cin >> a[i];
  }
  vector<llint> b(n + 2, 0);
  for (int i = 0; i <= n; i++) {
    b[i + 1] = b[i] + a[i];
  }

  llint now = 1, ans = 0;
  for (int i = 0; i <= n; i++) {
    llint c = now - a[i];
    if (c < 0) {
      cout << -1 << endl;
      return 0;
    }
    ans += a[i] + c;
    llint next = 2 * c;
    llint m = b.back() - b[i] - a[i];
    if (m < next) {
      next = m;
    }
    now = next;
  }

  cout << ans << endl;

  return 0;
}
