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<int64_t> p(N+1);
  int64_t X=0,Y=1,Z=0,A=0;
  for(int i=0;i<=N;i++){
    cin>>p.at(i);
    X+=p.at(i);
  }
  for(int i=0;i<=N;i++){
    if(Y<=0){
      A++;
      break;
    }
    if(A==0){
      Z+=min(X,Y);
    }
    else{
      Z+=X;
    }
    X-=p.at(i);
    if(Y>=100000000000000){
      A=2;
    }
    else{
      Y-=p.at(i);
      Y*=2;
    }
  }
  if(A==1||Y<0){
    cout<<"-1"<<endl;
  }
  else{
    cout<<Z<<endl;
  }
}