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++)
using ll = long long;

int main() {
  int n;
  cin >> n;
  vector<int64_t> vec(n+1);
  int64_t sum=0;
  bool b=true;
  rep(i,n+1){
    cin >> vec[i];
    sum+=vec[i];
  }
  int64_t now=1;
  int64_t ans=1;
  if(vec[0]>1)b=false;
  else if(vec[0]==1&&n!=0)b=false;
  else if(vec[0]==1) ans=1;
  else{
    rep(i,n){
      if(vec[i+1]>2*now) b=false;
      now=min(sum,2*now);
      ans+=now;
      sum-=vec[i+1];
      now-=vec[i+1];
      
    }
  }
  if(b==true) cout << ans << endl;
  else cout << -1 << endl;
    

}