D. Artsem and Saunders

Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.

Let [n] denote the set {1, ..., n}. We will also write f: [x] → [y] when a functionf is defined in integer points 1, ..., x, and all its values are integers from 1 toy.

Now then, you are given a function f: [n] → [n]. Your task is to find a positive integerm, and two functions g: [n] → [m],h: [m] → [n], such that g(h(x)) = x for all , andh(g(x)) = f(x) for all, or determine that finding these is impossible.

Input

The first line contains an integer n (1 ≤ n ≤ 10⁵).

The second line contains n space-separated integers — valuesf(1), ..., f(n) (1 ≤ f(i) ≤ n).

Output

If there is no answer, print one integer -1.

Otherwise, on the first line print the number m (1 ≤ m ≤ 10⁶). On the second line printn numbers g(1), ..., g(n). On the third line printm numbers h(1), ..., h(m).

If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.

Examples

Input

3
1 2 3

Output

3
1 2 3
1 2 3

Input

3
2 2 2

Output

1
1 1 1
2

Input

2
2 1

Output

-1
思路：由h(g(x)) = f(x),假设f(x1) = f(x2);

那么有h(g(x1)) = h(g(x2)),那么我们假设g(x1)!=g(x2),那么就有h中存在两个相等的数，那么g(h(g(x1))) = g(x1),g(h(g(x2))) = g(x2)
就有个g(x1) = g(x2)所以假设不成立，那么h中的数都是不一样的，那么就为f中的不同种类数，m就确定了。
于是确定h(x),那么h(x)有许多的排列，假设x1，x2，h(x1) = a,h(x2) = b,f(y1) = a,f(y2) = b;
通过条件1求出g,h(g(a)) = f(a),h(g(b)) = f(b);-> g(a) = x1,g(b) = x2,那么有h(x1) = f(a),h(x2) = f(b);
那么假设h(x1) = b,h(x2) = a,可以推得g(a) = x2,g(b) = x1,有h(x2) = f(a),h(x1) = f(b),总是可以的f（a）= a，f（b） = b;
所以h的序列任意排序即可，然后求出g，在通过第二个条件验证即可。

 1 #include<stdio.h> 2 #include<algorithm> 3 #include<queue> 4 #include<stdlib.h> 5 #include<iostream> 6 #include<string.h> 7 #include<set> 8 #include<map> 9 using namespace std;10 typedef long long LL;11 set<int>vec;12 set<int>::iterator it;13 int ans[1000006];14 int g[1000006];15 int h[1000006];16 map<int,int>my;17 int main(void)18 {19     int n;20     scanf("%d",&n);21     int cn = 0;22     for(int i = 1; i <= n; i++)23         scanf("%d",&ans[i]),vec.insert(ans[i]);24     for(it = vec.begin(); it!=vec.end(); it++)25     {26         h[++cn] = *it;27         my[*it] = cn;28         g[h[cn]] = cn;29     }30     int flag = 0;31     for(int i = 1; i <= n; i++)32     {33         if(g[i] == 0)34         {35             g[i] = my[ans[i]];36         }37         else if(h[g[i]]!=ans[i])38             flag = 1;39     }40     if(flag == 1)41         printf("-1\n");42     else43     {44         printf("%d\n",cn);45         printf("%d",g[1]);46         for(int i = 2; i <= n; i++)47             printf(" %d",g[i]);48         printf("\n");49         printf("%d",h[1]);50         for(int i = 2; i <= cn; i++)51             printf(" %d",h[i]);52     }53     return 0;54 }