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 sc = new Scanner(System.in);
		
		Map<Integer,Integer> map = new HashMap();
		
		int n = sc.nextInt();
		int f[] = new int[n],g[] = new int[n];
		List<Integer> h = new ArrayList();
		for(int i = 0 ;i<n ;i++){
			f[i] = sc.nextInt();
		}
		
		boolean b = true;
		int count=1;
		for(int i = 0 ;i<n ;i++){
			int temp = f[i];
			if(!map.containsKey(temp)){
				if(f[temp-1]!=temp){
					b=false;
					break;
				}
				else{
					h.add(temp);
					map.put(temp, count);
					g[i] = count;
					count++;
				}
			}
			else{
				g[i] = map.get(temp);
			}
		}
		
		if(b){
			System.out.println(h.size());
			for(int i : g)
				System.out.print(i+" ");
			System.out.println();
			for(int i : h)
				System.out.print(i+" ");
		}
		else
			System.out.print(-1);
	}
}