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>
#define int long long
using namespace std;

int N;
int suff[100005];
int A[100005];

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  int now = 1, ans = 0;
  for(int i = 0; i <= N; i++) {
    cin >> A[i];
  }
  for(int i = N; i >= 0; i--) {
    suff[i] += suff[i + 1] + A[i];
  }
  for(int i = 0; i <= N; i++) {
    cin >> A[i];
    ans += min(now, suff[i]);
    now -= A[i];
    now *= 2;
    if(now < 0) {
      cout << "-1\n";
      return 0;
    }else if(now > 1e18) now = 1e18;
  }
  cout << ans << "\n";
}
