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 <iostream>
#include <vector>
#include <bitset>
#include <utility>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
 
using namespace std;
typedef long long ll;
 
int main(){
  int N;
  cin >> N;
  ll sum=0;
  vector<ll> A(N+1);
  for (int i = 0; i < N+1; i++){
    cin >> A[i]; sum+=A[i];
  }
  
  ll ans=0;
  ll n=1;
  for(int i=0;i<N+1;i++){
    if(A[i]<=n){
      ans+=n;
      n-=A[i];
      sum-=A[i];
      n=min(2*n,sum);
    }else{
      cout << -1 << endl;
      return 0;
    }
  }
  cout << ans << endl;

  return 0;
}
