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(void) {
  int N; cin >> N;
  int A[N+1];
  long long child = 0;
  for (int i = 0; i <= N; i++) {
    cin >> A[i];
    child += A[i];
  }

  long long ans = 0, node = 1, cnum = child;

  for (int i = 0; i <= N; i++) {
    ans += node;
    cnum -= A[i];
    node -= A[i];

    if (i != N) {
      if (node <= 0) {
        cout << -1 << endl;
        return 0;
      }
    } else {
      if (node < 0) {
        cout << -1 << endl;
        return 0;
      }
    }

    node = min(node*2, cnum);
  }

  cout << ans << endl;

  return 0;
}