Codeforces 741A Arpa's loud Owf and Mehrdad's evil plan(思维)

 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.

题意：给出n个整数， 转换顺序为 x→a[x]→a[a[x]]→.... 一次规律继续进行下去t次，若终点为y，且满足

 y→a[y]→a[a[y]]→... 也进行t次 得到终点为x 。就称之为一个环。 要将数组里面分成若干个环，求最小的t为多

大，不存在输出-1。

题解：每一个环的长度为len ，满足任意 len * m(正整数) = t 。所以若t存在，那么t是所有环的长度的最小公倍

数。 如果环一定存在，那么每个x都可以按照上面的规则回到自己的位置，自环的长度为 temp，如果temp为奇

数，那么这个自环就是我们要找的环； 如果temp为偶数，那么 只要 temp/2的长度 x就可以到达y，y也可以到达

x，所以环长度为 temp/2。 

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 105;int a[maxn],mark[maxn];LL gcd(LL a,LL b){if(b>a)swap(a,b);LL c;while(b>0){c=a%b;a=b;b=c;}return a;}int main(){int n;while(scanf("%d",&n)!=EOF){memset(mark,0,sizeof(mark));for(int i=1;i<=n;++i)scanf("%d",&a[i]);LL ans=1;int flag=0;for(int i=1;i<=n;++i){if(mark[i])continue;LL temp=1;int y=a[i];mark[i]=1;while(y!=i){mark[y]=1;y=a[y];temp++;if(temp>n){//环不存在 flag=1;break;}}if(flag)break;if(temp%2==0)temp/=2;ans = ans/gcd(ans,temp)*temp;//求最小公倍数 }if(flag)printf("-1\n");else printf("%lld\n",ans);}return 0;}