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:
// g++ -std=c++11 a.cpp
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>   
#include<map>
#include<set>
#include<unordered_map>
#include<utility>
#include<cmath>
#include<random>
#include<cstring>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<sstream>
#include<iomanip>
#include<assert.h>
#include<typeinfo>
#define loop(i,a,b) for(ll i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define FOR(i,a) for(auto i:a)
#define pb push_back
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
#define show1d(v) rep(i,v.size())cout<<" "<<v[i];cout<<endl<<endl;
#define show2d(v) rep(i,v.size()){rep(j,v[i].size())cout<<" "<<v[i][j];cout<<endl;}cout<<endl;
using namespace std;
//kaewasuretyuui
typedef long long ll;
//#define int ll
typedef int Def;
typedef pair<Def,Def> pii;
typedef vector<Def> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<Def,pii> pip;
typedef vector<pip>vip;
#define mt make_tuple
typedef tuple<int,int,int,int> tp;
typedef vector<tp> vt;
template<typename A,typename B>bool cmin(A &a,const B &b){return a>b?(a=b,true):false;}
template<typename A,typename B>bool cmax(A &a,const B &b){return a<b?(a=b,true):false;}
//template<class C>constexpr int size(const C &c){return (int)c.size();}
//template<class T,size_t N> constexpr int size(const T (&xs)[N])noexcept{return (int)N;}
const double PI=acos(-1);
const double EPS=1e-9;
Def inf = sizeof(Def) == sizeof(long long) ? 2e18 : 1e9+10;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
int n;
vi in,to;
ll vtol(vi a){
	ll t=0;
	rep(i,a.size()){
		t*=10;
		t+=a[i];
	}
	return t;
}
map<ll,int>tmp;
void f(vi a,int c){
	if(c==(n-1)/2)return;
	rep(i,n-1)loop(j,i+1,n){
		vi t=a;
		loop(k,i,j+1)if(k<j-k+i)
			swap(t[k],t[j-k+i]);
		ll s=vtol(t);
		if(tmp.count(s))tmp[s]=min(tmp[s],c+1);
		else tmp[s]=c+1;
		f(t,c+1);
	}
}
int main(){
	cin>>n;
	in=vi(n);
	rep(i,n)cin>>in[i];
	to=vi(n);
	rep(i,n)to[i]=i,in[i]--;
	tmp[vtol(to)]=0;
	f(to,0);
	map<ll,int>A=tmp;
	tmp.clear();
	tmp[vtol(in)]=0;
	f(in,0);
	map<ll,int>B=tmp;
	int out=n-1;
	for(auto it:A)if(B.count(it.first)){
		out=min(out,A[it.first]+B[it.first]);
	}
	cout<<out<<endl;
}