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 Main {
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int tot=0;
		int cnt=0;
		int i=0;
		int pp=0;
		int[] a=new int[100010];
		int[] b=new int[100010];
		int[] c=new int[100010];
		for(i=1;i<=n;i++)
		{
			a[i]=in.nextInt();
			if(a[i]>n||a[i]<1)break;
			if(b[a[i]]==0) {
				b[a[i]]=1;
				tot++;
			}
			if(a[i]==i)cnt++;
		}
		if(i<=n||cnt!=tot)System.out.println("-1");
		else {
			System.out.printf("%d\n",tot);
			for(i=1;i<=n;i++)
			if(b[i]==1) {
				pp++;
				b[i]=pp;
				c[pp]=i;
			}
			for(i=1;i<=n;i++)
				System.out.printf("%d ",b[a[i]]);
			System.out.println();
			for(i=1;i<=pp;i++)
				System.out.printf("%d ",c[i]);
			System.out.println();
		}
		in.close();
	}
}
