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 function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.

Now then, you are given a function f: [n] → [n]. Your task is to find a positive integer m, and two functions g: [n] → [m], h: [m] → [n], such that g(h(x)) = x for all <image>, and h(g(x)) = f(x) for all <image>, 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 — values f(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 print n numbers g(1), ..., g(n). On the third line print m 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
import java.util.*;

public class CF397D {
	public static void main(String []args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int []f=new int[n+1];
		int m=0;
		int []hash=new int[n+1];
		int []h=new int[n+1];
		int []g=new int[n+1];
		for(int i=1;i<=n;i++)
			f[i]=sc.nextInt();
		for(int i=1;i<=n;i++){
			if(f[f[i]]!=f[i]){
				System.out.println(-1);
				return;
			}		
		}
		for(int i=1;i<=n;i++){
			if(hash[f[i]]==0){
				m+=1;
				hash[f[i]]=m;
				h[m]=f[i];
			}
			g[i]=hash[f[i]];
		}
		System.out.println(m);
		for(int i=1;i<=n;i++)
			System.out.print(g[i]+" ");
		System.out.println();
		for(int i=1;i<=m;i++)
			System.out.print(h[i]+" ");
		System.out.println();
		
	}
}
