Problem: A permutation of N numbers from 1 to N A1, A2, ……, AN is given. You can perform the operation reverse (i, j) to reverse the order of the numbers in the interval [i, j] (1 ≤ i ≤ j ≤ N) for this permutation. For example, if reverse (2, 4) is applied to [1, 2, 3, 4, 5], it becomes [1, 4, 3, 2, 5]. Calculate how many operations you need to do at a minimum to get the permutation sorted in ascending order.

Constraints

* N is an integer

* 2 ≤ N ≤ 10

Input

The input is given in the following format.

> N
> A1 A2 …… AN
>

Output

Print the solution to the problem on one line.

Examples

Input

5
1 4 3 5 2


Output

2


Input

5
3 1 5 2 4


Output

4


Input

3
1 2 3


Output

0


Input

10
3 1 5 2 7 4 9 6 10 8


Output

9

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
map<vector<int>,int> bfs(vector<int> A){
  map<vector<int>,int> D;
  queue<vector<int> > Q;
  Q.push(A);
  D[A] = 0;
  while(!Q.empty()){
    vector<int> t = Q.front();Q.pop();
    int cost = D[t];
    if(cost == 4) continue;
    
    for(int i=0;i<n;i++)
      for(int j=i+2;j<=n;j++){
	vector<int> nt = t;
	reverse(nt.begin()+i,nt.begin()+j);
	if(D.count(nt)) continue;
	Q.push(nt);
	D[nt]=cost+1;
    }
  }
  return D;
}


int main(){
  cin>>n;
  vector<int> A(n),B(n);
  for(int i=0;i<n;i++) cin>>A[i],B[i] = i+1;

  
  map<vector<int>,int> X = bfs(A);
  map<vector<int>,int> Y = bfs(B);

  map<vector<int>,int>::iterator it = X.begin();
  int ans = 9;
  
  for(;it!=X.end();it++)
    if(Y.count(it->first))ans = min(ans,it->second+Y[it->first]);
  
  cout<<ans<<endl;
  


  return 0;
}