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 <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<long long> vll;
typedef pair<int,int> pint;
typedef pair<long long, long long> pll;

#define MP make_pair
#define PB push_back
#define ALL(s) (s).begin(),(s).end()
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P) 
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P) 
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P) 
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s; }



int N;
map<vint,int> back;
map<vint,int> dp;

int solve() {
    vint target(N);
    vint start(N);
    for (int i = 0; i < N; ++i) {
        target[i] = i+1;
        cin >> start[i];
    }
    queue<vint> que;
    que.push(target);
    back.clear();
    back[target] = 0;
    
    while (!que.empty()) {
        vint now = que.front(); que.pop();
        //cout << now << " : " << back[now] << endl;
        if (back[now] == 4) continue;
        if (now == start) { return back[now]; }
        
        for (int i = 0; i < N; ++i)
            for (int j = i+2; j <= N; ++j) {
                vint next = now;
                reverse(next.begin() + i, next.begin() + j);
//                for (int k = 0; k < i; ++k) next[k] = now[k];
//                for (int k = i; k < j; ++k) next[k] = now[i+j-1 - k];
//                for (int k = j; k < N; ++k) next[k] = now[k];
                
                if (!back.count(next)) {
                    back[next] = back[now] + 1;
                    que.push(next);
                }
            }
    }
    
    que.push(start);
    dp.clear();
    dp[start] = 0;
    
    while (!que.empty()) {
        vint now = que.front(); que.pop();
        //cout << now << " : " << dp[now] << endl;
        if (dp[now] == 4) continue;
        if (back.count(now)) return dp[now] + back[now];
            
        for (int i = 0; i < N; ++i)
            for (int j = i+1; j <= N; ++j) {
                vint next = now;
                reverse(next.begin() + i, next.begin() + j);
                    
                if (!dp.count(next)) {
                    dp[next] = dp[now] + 1;
                    que.push(next);
                }
            }
    }
    
    return 9;
}

int main() {
    while (cin >> N) {
        cout << solve() << endl;
    }
    return 0;
}