//第一种：使用string.h中的strrev函数，只用于字符串
#include <iostream>
#include <cstring>
using namespace std;
int main()
{  char s[]="hello";  strrev(s);  cout<<s<<endl;   return 0;
}  //第二种：使用algorithm中的reverse函数，也适用于整型数组及其他存储结构
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;  int main()
{  string s="hello";  reverse(s.begin(),s.end());  cout<<s<<endl;  return 0;
}   

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define mod1 2333333
#define mod2 1048577
#define maxn 51
using namespace std;
int n,t[maxn];
struct node{int a[maxn],step;
}cur,nxt;
queue<node>q;
bool vis1[2333333],vis2[1048577];
int hash1(int x[]){int res=1;for(int i=1;i<=n;i++)res=(res*103%mod1+x[i])*13%mod1;return res;
}
int hash2(int x[]){int res=1;for(int i=1;i<=n;i++)res=(res*117%mod2+x[i])*27%mod2;return res;
}
int main(){freopen("Cola.txt","r",stdin);scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&cur.a[i]),t[i]=cur.a[i];sort(t+1,t+n+1);int t1=hash1(t),t2=hash2(t);if(hash1(cur.a)==t1&&hash2(cur.a)==t2){puts("0");return 0;}cur.step=0;q.push(cur);while(!q.empty()){cur=q.front();q.pop();for(int i=2;i<=n;i++){nxt=cur;reverse(nxt.a+1,nxt.a+i+1);int h1=hash1(nxt.a),h2=hash2(nxt.a);if(h1==t1&&h2==t2){printf("%d",cur.step+1);return 0;}if(!vis1[h1]||!vis2[h2]){vis1[h1]=1;vis2[h2]=1;nxt.step=cur.step+1;q.push(nxt);}}}return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define maxn 51
using namespace std;
int n,a[maxn];
struct node{int w,id;bool operator < (const node b)const{return w<b.w;}
}e[maxn];
int get(){int res=0;for(int i=2;i<=n;i++)if(abs(a[i]-a[i-1])!=1)res++;return res;
}
void dfs(int limit,int step){int h=get();if(!h&&a[1]<a[2]){printf("%d",limit);exit(0);}if(h+step>limit||step==limit)return;for(int i=2;i<=n;i++){if(abs(a[i]-a[i+1])!=1){reverse(a+1,a+i+1);dfs(limit,step+1);reverse(a+1,a+i+1);}}
}
int main(){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&e[i].w),e[i].id=i;sort(e+1,e+n+1);for(int i=1;i<=n;i++)a[e[i].id]=i;a[0]=-0x7f7f7f7f;a[n+1]=0x7f7f7f7f;for(int k=0;;k++)dfs(k,0);
}