As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if xstarts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
4 2 3 1 4
3
4 4 4 4 4
-1
4 2 1 4 3
1
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
由题意 其需满足条件 所有的点必须在一个环中，包括为1的环
否则输出-1
环长度为奇数，长度不变，如果为偶数，则除以二，为0改为1
求所有环长度的最小公倍数
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
ll n;
int a[105];
int flag[105];
int main()
{
ios::sync_with_stdio(false);
ll re = -1;
ll t;
cin >> n;
for(int i=1;i<=n;i++)
cin >> a[i];
for(int i=1;i<=n;i++)
{
if(flag[i])
continue;
t = 0;
ll cur = a[a[i]];
flag[a[i]] = 1;
flag[cur] = 1;
t++;
while(cur!=a[i])
{
cur = a[cur];
flag[cur] = 1;
t++;
if(t>n)
{
cout << "-1" <<endl;
return 0;
}
}
if(t%2==0)//根据环数奇偶判断
t/=2;
if(t==0)
t=1;
if(re==-1)
{
re = t;
}
else
re = (re*t)/gcd(re,t);//求最小公倍数
//cout << t <<endl;
}
for(int i=1;i<=n;i++)
if(flag[i]==0)
{
cout << "-1" <<endl;
return 0;
}
cout << re <<endl;
return 0;
}