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>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

using ll=long long;
using R=long double;
const R EPS=1e-9; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max<R>(x,0.0));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

// Problem Specific Parameter:

int ans=9;
map<vector<int>,int> memo;

void make(vector<int> &a,int d){
	const int n=a.size();
	
	if(memo.find(a)==end(memo)) memo[a]=d;
	chmin(memo[a],d);

	if(d==4) return;
	rep(j,n)rep(i,j){
		reverse(begin(a)+i,begin(a)+j+1);
		make(a,d+1);
		reverse(begin(a)+i,begin(a)+j+1);
	}
}


void dfs(vector<int> &a,int d){
	const int n=a.size();
	if(memo.find(a)!=end(memo)) chmin(ans,d+memo[a]);

	if(d==4) return;
	rep(j,n)rep(i,j){
		reverse(begin(a)+i,begin(a)+j+1);
		dfs(a,d+1);
		reverse(begin(a)+i,begin(a)+j+1);
	}
}



int main(void){
	int n;
	cin >> n;

	vector<int> in(n);	
	rep(i,n) cin >> in[i];
	make(in,0);

	vector<int> init(n);
	iota(begin(init),end(init),1);
	dfs(init,0);

	cout << ans << endl;
	return 0;
}