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:
//
//  main.cpp
//  AOJ
//
//  Created by user on 2014/06/01.
//  Copyright (c) 2014年 user. All rights reserved.
//

#include <iostream>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <cstring>
using namespace std;

const int SIZE = 3628810;
typedef pair< vector<int>, int > Pack;

int getPermId(const vector<int>& v){
    const int N = (int)v.size();
    
    int fact = 1;
    int res = 0;
    
    for(int i=N-1;i>=0;i--){
        for(int j=i+1;j<N;j++){
            if(v[j] < v[i]){
                res += fact;
            }
        }
        fact *= (N - i);
    }
    
    return res;
}

int dist[2][SIZE];
vector<Pack> Queue[2];

int bfs(const vector<int>& start, const vector<int>& goal){
    memset(dist, -1, sizeof(dist));
    
    const int N = (int)start.size();
    const int startId = getPermId(start);
    const int goalId  = getPermId(goal);
    
    if(startId == goalId){
        return 0;
    }
    
    int curr = 0, next = 1;
    
    dist[curr][startId] = 0;
    dist[next][goalId]  = 0;
    
    Queue[curr].push_back(make_pair(start, startId));
    Queue[next].push_back(make_pair(goal,  goalId));
    
    for(;;){
        vector<Pack> nextQueue;
        for(int i=0;i<Queue[curr].size();i++){
            Pack &pack = Queue[curr][i];
            
            vector<int> &perm = pack.first;
            int permId = pack.second;
            int nowCost = dist[curr][permId];
            
            for(int s=0;s<N;s++)for(int t=s+1;t<N;t++){
                reverse(perm.begin() + s, perm.begin() + t + 1);
            
                int nextPermId = getPermId(perm);
                if(dist[curr][nextPermId] == -1){
                    dist[curr][nextPermId] = nowCost + 1;
                
                    if(dist[next][nextPermId] != -1){
                        return dist[curr][nextPermId] + dist[next][nextPermId];
                    } else {
                        nextQueue.push_back(make_pair(perm, nextPermId));
                    }
                }
                
                reverse(perm.begin() + s, perm.begin() + t + 1);
            }
        }
        Queue[curr].swap(nextQueue);
        
        swap(curr, next);
    }
    
    return 1;
}

int main(){
    int N;
    cin >> N;
    
    vector<int> start(N);
    for(int i=0;i<N;i++){
        cin >> start[i];
        --start[i];
    }
    
    vector<int> goal = start;
    sort(goal.begin(), goal.end());
    
    int res = bfs(start, goal);
    cout << res << endl;
    
    return 0;
}