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;
  long long a[n+1],sum=0,now=1,ans=1;
  for(int i=0;i<n+1;i++){
    long long l;
    cin>>l;
    a[i]=l;
    sum+=a[i];
  }
  bool ok=true;
  if(a[0]!=0)ok=false;
  if(a[0]==1&&n==0)ok=true;
  for(int i=1;i<=n;i++){
    if(2*now<a[i]){
      ok=false;
      break;
    }
    if(sum<now*2){
      now=sum;
    }
    else{
      now*=2;
    }
    ans+=now;
    now-=a[i];
    sum-=a[i];
  }
  if(ok)cout<<ans<<endl;
  else cout<<-1<<endl;
}