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>
#define rep(i,n) for(int i = 0; i < n; i++)
#define pb push_back
using namespace std;
typedef long long ll;

int main(){
  int n;
  cin>>n; 
  vector<ll> a(n+1);
  rep(i,n+1) cin>>a[i];
  
  vector<ll> b(n+1);
  
  vector<ll> c(n+1);
  c[n]=a[n];
  for(int i=n;i>0;i--) c[i-1]=c[i]+a[i-1];
  
  b[0]=1;
  rep(i,n){
    if(b[i]-a[i]<=0){
      cout<<-1<<endl;
      return 0;
    }
    b[i+1]=min((b[i]-a[i])*2,c[i+1]);
  } 
  if(b[n]!=a[n]){
    cout<<-1<<endl;
    return 0;
  }
  
  ll ans=0;
  rep(i,n+1){
    ans+=b[i];
  }
  cout<<ans<<endl;
}