A. Arpa's loud Owf and Mehrdad's evil plan－－java实现

A. Arpa's loud Owf and Mehrdad's evil plan

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ttimes) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1times) 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 x starts 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).

Input

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.

Output

If there is no t satisfying the condition, print -1. Otherwise print such smallest t.

Examples

input

42 3 1 4

output

3

input

44 4 4 4

output

-1

input

42 1 4 3

output

1

Note

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.

n个整数，x→a[x]→a[a[x]]→.... 进行t次，“owwwf”可以不看＝ ＝，当终点为y，且y进行t次也能得到终点为x 则在数组里存在一个环。 求数组中环的最小公倍数，不存在输出-1。其实可以看作求这些数组中环长度的最小公倍数。通俗点讲按第一个例子来，当x＝2，a［2］＝3，注意这里数组的a［0］不使用，实际程序自己可以处理，此时x2＝3，a［3］＝1，此时x3=1，a［1］＝x＝2，形成一个环。写的有点粗糙，也借鉴了别人的思路进行了改进。

import java.util.*;public class M{    static Scanner sc;    public static void swap(long a, long b){ long tmp = a;a = b;b = tmp;    }    public static long gcd(long a,long b)    {if(b>a)    swap(a,b);        long c;while(b>0){    c=a%b;   a=b;   b=c;}return a;    }    public static void main(String []args){        sc = new Scanner(System.in);        int n=sc.nextInt();        int num=1;        int []a=new int[n+1];        int []b=new int[n+1];        while (sc.hasNext()){            a[num]=sc.nextInt();            num++;        }        long ans=1;int flag=0;for(int i=1;i<=n;++i){if(b[i]==1)continue;long temp=1;int y=a[i];b[i]=1;while(y!=i){b[y]=1;y=a[y];temp++;if(temp>n){ flag=1;break;}}if(flag==1)break;if(temp%2==0)temp/=2;ans = ans/gcd(ans,temp)*temp; }if(flag==1)System.out.print("-1\n");else System.out.print(ans);    }}