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;
  n++;
  vector<int> a(n);
  vector<int64_t> s(n,0);
  for(int i=0;i<n;i++){
    cin>>a.at(i);
    s.at(0)+=a.at(i);
  }
  for(int i=1;i<n;i++)
    s.at(i)=s.at(i-1)-a.at(i-1);
  vector<int64_t> c(n);
  c.at(0)=1;
  int64_t x=1-(a.at(0));
  bool f=1;
  for(int i=1;i<n;i++){
    x=min(x*2,s.at(i));
    if(x>=a.at(i)){
      c.at(i)=x;
      x-=a.at(i);
    }else{
      f=0;
      break;
    }
  }
  if(x==0&&f){
    int64_t v=0;
    for(int i=0;i<n;i++)
      v+=c.at(i);
    cout<<v<<endl;
  }else
    cout<<-1<<endl;
}