Codeforces Round #383 (Div. 2)C. Arpa's loud Owf and Mehrdad's evil plan

C. 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 crush_i.

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 crush_x and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crush_x calls crush_crushx 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 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. crush_i = 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 crush_i (1 ≤ crush_i ≤ 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

4
2 3 1 4

Output

3

Input

4
4 4 4 4

Output

-1

Input

4
2 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.

/*
并查集瞎搞，唯一一组爆int的数据：
如果环的长度为：2 3 5 7 11 13 17 19 23，那么最小公倍数就是363322674就会爆int
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int bin[];
int vis[];
int vis2[];
int n,a[];
ll cur=;
ll gcd(ll a,ll b)
{
    return a==?b:gcd(b%a,a);
}
ll lcm(ll a,ll b)
{
    return a*b/gcd(a,b);
}
int findx(int x,int start)//查找
{
    while(bin[x]!=x)
    {
        x=bin[x];
        if(x==start)
            break;
    }
    return x==start;
}
void op(int x,int start)//查找
{
    vis[x]=;
    int flag=;//表示是不是第一次
    while(bin[x]!=x)
    {

        vis[x]=;
        x=bin[x];
        if(x==start)
        {
            if(cur!=)
                cur++;
            break;
        }
        cur++;
    }
    if(cur==)
        cur=;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=;i<=n;i++)
    {
        vis[i]=;
        vis2[i]=;
        bin[i]=i;
    }
    int ans=;
    for(int i=;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(vis2[a[i]]==)
            ans++;
        bin[i]=a[i];
        vis2[a[i]]=;
    }
    if(ans!=n)
    {
        puts("-1");
        return ;
    }
    int f=;//是不是每个节点都有环
    long long maxn=;
    for(int i=;i<=n;i++)
    {
        cur=;
        //cout<<"findx(i,i)="<<findx(i,i)<<endl;
        if(vis[i]==)//没有访问过
        {
            if(findx(i,i))//有环
            {
                op(i,i);//去更新操作
                //cout<<"cur="<<cur<<endl;
                if(cur%==)
                    cur/=;
                maxn=maxn/__gcd(maxn,cur)*cur;
            }
            else
            {
                f=;
                break;
            }
        }
    }
    if(f)
        puts("-1");
    else
        printf("%lld\n",maxn);
    return ;
}