Problem: 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

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

/**
 * Created by Anton Berezin on 2/14/2017.
 */
public class Sanders {
	public static void main(String[] args) throws Exception {
		try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) {

			int n = Integer.parseInt(reader.readLine());
			StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
			int[] f = new int[n + 1];
			Set<Integer> distinctFs = new HashSet<>();
			for (int i = 1; i <= n; i++) {
				f[i] = Integer.parseInt(tokenizer.nextToken());
				distinctFs.add(f[i]);
			}
			int m = distinctFs.size();
			Map<Integer, Integer> dist = new HashMap<>();
			int distNum = 1;
			for (int i = 1; i <= n; i++) {
				if (!dist.containsKey(f[i])) {
					dist.put(f[i], distNum);
					distNum++;
				}
			}
			int[] g = new int[n + 1];
			for (int i = 1; i <= n; i++) {
				g[i] = dist.get(f[i]);
			}
			int[] h = new int[m + 1];

			for (int i = 1; i <= n; i++) {
				h[g[i]] = f[i];
			}

			boolean can = true;
			for (int i = 1; i <= m; i++) {
				if (g[h[i]] != i) {
					can = false;
					break;
				}
			}
			if (!can) {
				System.out.println(-1);
				return;
			}
			System.out.println(m);
			StringBuilder builder = new StringBuilder();
			for (int i = 1; i <= n; i++) {
				builder.append(g[i]).append(" ");
			}
			System.out.println(builder.toString().trim());
			builder = new StringBuilder();
			for (int i = 1; i <= m; i++) {
				builder.append(h[i]).append(" ");
			}
			System.out.println(builder.toString().trim());
		}
	}
}