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 思路：每个点开始，因为x->y,y->x,那么这必定是一个环，所以就先找到环，然后假设环上不同的两点，相距x1，L-x1; 那么从一点开始到达另一点走x1+k1*L,那么另一点就为L - x1 + k2*L ; 这两个要相同，那么有X%L = x1&&X%L = L-x1;所以当L为偶数时最小为X=L/2;否则为X=L; 最后每个环求最小公倍数。

1 #include
     
       2 #include
      
        3 #include
       
         4 #include
        
          5 #include
         
           6 #include
          
            7 #include
           
             8 #include
            
              9 #include
             
              10 typedef long long LL;11 using namespace std;12 int ans[1005];13 LL cnt[1000005];14 int ma[105];15 bool flag[105];16 LL gcd(LL n,LL m);17 int main(void)18 {19 int n;20 scanf("%d",&n);21 int i,j;22 for(i = 1; i <= n; i++)23 {24 scanf("%d",&ans[i]);25 }26 memset(ma,-1,sizeof(ma));27 int id = -1;28 for(i = 1; i <= n; i++)29 {30 int t = 0;31 memset(flag,0,sizeof(flag));32 int v = ans[i];33 while(!flag[v])34 {35 t++;36 flag[v] = true;37 cnt[v] = t;38 v=ans[v];39 if(v == i)40 {t++;ma[i] = t;break;}41 }42 }43 int fl = 0;LL ak = 1;44 for(i = 1; i <=n; i++)45 {46 if(ma[i]==-1)47 fl = 1;48 else49 {50 if(ma[i]%2==0)51 ma[i]/=2;52 LL ac = gcd(ma[i],ak);53 ak = ak/ac*ma[i];54 }55 }56 if(fl)printf("-1\n");57 else printf("%lld\n",ak);58 return 0;59 }60 LL gcd(LL n,LL m)61 {62 if(m == 0)63 return n;64 else return gcd(m,n%m);65 }