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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main(){
  int N; cin >> N;
  long long num = 1;
  long long ans = 0;

  vector<int> A(N + 1); rep(i, N + 1) cin >> A[i];
  long long sum = 0;
  rep(i, N + 1) sum += A[i];
  vector<long long> rA(N + 2);
  rep(i, N + 2) {
    rA[i] = sum;
    sum -= A[i];
  }

  rep(i, N + 1) {
    if (A[i] > num) {
      ans = -1;
      break;
    }
    ans += num;
    num -= A[i];
    num = min(rA[i + 1], num * 2);
  }
  cout << ans << endl;
}
