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() {
  int N;
  cin>>N;
  vector<int> A(N+1);
  for(int &i:A)
    cin>>i;
  bool bo=false;
  for(int i=0,j=1;i<=N;j*=2){
    j-=A.at(i);
    i++;
    if(j<0){
      bo=true;
      break;
    }else if(j>=200000000)
      j=200000000;
  }
  if(bo){
    cout<<-1<<endl;
    return 0;
  }
  int64_t s=0;
  for(int &i:A)
    s+=i;
  int64_t n=1,k=0;
  for(int i=0;i<=N;i++){
    k+=n;
    n-=A.at(i);
    s-=A.at(i);
    if(n*2<=s)
      n*=2;
    else if(n*2>s && n<s)
      n=s;
  }
  cout<<k<<endl;
}