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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner cin = new Scanner(System.in);
		int n = cin.nextInt();
		int f[] = new int[n + 1];
		int vis[] = new int[n + 1];
		for(int i = 1 ; i <= n ; i ++){
			f[i] = cin.nextInt();
		}
		
		int h[] = new int[1000010];
		int g[] = new int[n + 1];
		int m = 0;
		for(int i = 1 ; i <= n ; i ++){
			if(vis[f[i]] == 0){
				m ++;
				h[m] = f[i];
				vis[f[i]] = m;
			}
			g[i] = vis[f[i]];
		}
		boolean flag = true;
		for(int i = 1 ; i <= m ; i ++){
			if(g[h[i]] != i){
				flag = false;
				break;
			}
		}
		if(!flag)System.out.println("-1");
		else {
			System.out.println(m);
			for(int i = 1 ; i <=n ; i++){
				if(i == 1)
					System.out.print(g[i]);
				else 
					System.out.print(" "+g[i]);
			}
			System.out.println("");
			for(int i = 1 ; i <= m ; i ++){
				if(i == 1)System.out.print(h[i]);
				else System.out.print(" "+ h[i]);
			}
		}
	}

}