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() {
  long n,aaa=-1,bbb=1,ans=0;
  cin >> n;
  vector<long> a(n+1);
  for(long i=0;i<=n;i++) cin >> a[i];
  for(long i=0;i<=n;i++) {
    aaa+=a[i];
  }

  for(long i=0;i<=n;i++) {
    ans+=bbb;
    bbb-=a[i];
    if(bbb<0) {
      ans=-1;break;
    }
    long x=min(aaa,bbb);
    bbb+=x;
    aaa-=x;
  }
  
  cout << ans << endl;
}