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 <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <climits>
#include <cstring>
#include <cassert>


using namespace std;
#define SZ(x) (int)(x.size())
#define REP(i, n) for(int i=0;i<n;++i)
#define FOR(i, a, b) for(int i=a;i<b;++i)
#define RREP(i, n) for(int i=n-1;i>=0;--i)
#define RFOR(i, a, b) for(int i=b-1;i>=a;--i)

#define all(x) begin(x),end(x)
#define dump(x) cerr<<#x<<" = "<<(x)<<endl
#define debug(x) cerr<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<< endl;

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using P = pair<int, int>;
const double eps = 1e-8;
const ll MOD = 1000000007;
const int INF = INT_MAX / 2;
const ll LINF = LLONG_MAX / 2;

template<class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}


template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> p) {
    os << p.first << ":" << p.second;
    return os;
}

template<class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
    os << "[";
    REP(i, SZ(v)) {
        if (i) os << ", ";
        os << v[i];
    }
    return os << "]";
}

template<class T1, class T2>
ostream &operator<<(ostream &os, const map<T1, T2> &mp) {
    os << "[";
    for (auto it = mp.begin(); it != mp.end(); ++it) {
        if (it != mp.begin()) {
            os << ", ";
        }
        os << *it;
    }
    return os << "]";
}

// edit
int N;
using Int = long long; // hash に使う
Int base = 10; // hash に使う
int dist_lim = 4; //距離がこれ以内を全列挙

Int hash_vec(const vector<int> &A) {
    Int ret = 0ll;
    for (auto e : A) {
        ret *= base;
        ret += e;
    }
    return ret;
}

// [l, r]をreverseした場合
Int next_hash(Int h, int l, int r) {
    vector<int> A;
    for (int i = 0; i < N; ++i) {
        A.push_back(h % base);
        h /= base;
    }
    reverse(all(A));

    reverse(A.begin() + l, A.begin() + r + 1);
    Int new_hash = hash_vec(A);
    return new_hash;
}

// get shortest distance
void get_sd(map<Int, int> &ret, Int start_hash) {
    queue<Int> que;
    que.push(start_hash);
    ret[start_hash] = 0;

    while (!que.empty()) {
        Int top = que.front();
        int dist = ret[top];
        que.pop();
        if (dist >= dist_lim) continue;

        for (int l = 0; l < N; ++l) {
            for (int r = l + 1; r < N; ++r) {
                Int next = next_hash(top, l, r);
                if (!ret.count(next)) {
                    ret[next] = dist + 1;
                    que.push(next);
                } else {
//                    if (ret[next] > dist + 1) {
//                        ret[next] = dist + 1;
//                        que.push(next);
//                    }
//                    chmin(ret[next], dist + 1);
//                    ret[next] = min(ret[next], dist + 1);
                }
            }
        }
    }
}

void solve() {
    cin >> N;
    vector<int> A(N);
    REP(i, N) cin >> A[i];
    REP(i, N) A[i]--;

    map<Int, int> S, G; // key : hash, value : dist
    vector<int> start_vec;
    REP(i, N) start_vec.push_back(i);
    Int start_hash = hash_vec(start_vec);
    Int goal_hash = hash_vec(A);

    get_sd(S, start_hash);
    get_sd(G, goal_hash);
    int ans = N - 1;

    for (auto p : S) {
        Int h = p.first;
        int d = p.second;

        if (G.count(h)) {
            int dist = d + G[h];
//            chmin(ans, dist);
            if (ans > dist) {
                ans = dist;
                continue;
            }
        }


//        for (int l = 0; l < N; ++l) {
//            for (int r = l + 1; r < N; ++r) {
//                Int next = next_hash(h, l, r);
//                if (G.count(next)) {
//                    int dist = d + G[next] + 1;
//                    if (ans > dist) {
//                        ans = dist;
//                    }
////                    chmin(ans, dist);
//                }
//            }
//        }

    }

    cout << ans << endl;

//    {
//        queue<Int> sque;
//        auto vec = vector<int>();
//        for (int i = 1; i <= N; ++i) vec.push_back(i);
//        Int s = hash_vec(vec);
//        sque.push(s);
//
//
//    }


}


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
//    std::ifstream in("input.txt");
//    std::cin.rdbuf(in.rdbuf());

    solve();


    return 0;
}