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(){
  int64_t N;
  cin >> N;
  vector<int64_t> vec(N+1);
  int64_t sum = 0;
  for(int i=0; i<N+1; i++){
    cin >> vec[i];
    sum += vec[i];
  }
  
  bool flag = true;
  int64_t ans = 0;
  int64_t bottom = 1;
  for(int i=0; i<N+1; i++){
    bottom -= vec[i];
    if(bottom < 0){
      flag = false;
    }
    ans += bottom;
    ans += vec[i];
    sum -= vec[i];
    if(bottom < sum){
      bottom = min(bottom*2,sum);
    }
  }
  if(!flag){
    ans = -1;
  }
  cout << ans << endl;
}