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
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF765D extends PrintWriter {
	CF765D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF765D o = new CF765D(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int[] ff = new int[n + 1];
		for (int i = 1; i <= n; i++)
			ff[i] = sc.nextInt();
		for (int i = 1; i <= n; i++) {
			int f = ff[i];
			if (ff[f] != f) {
				println(-1);
				return;
			}
		}
		int[] cc = new int[n + 1];
		int[] hh = new int[n + 1];
		int m = 0;
		for (int i = 1; i <= n; i++)
			if (ff[i] == i) {
				hh[++m] = i;
				cc[i] = m;
			}
		int[] gg = new int[n + 1];
		for (int i = 1; i <= n; i++) {
			int f = ff[i];
			gg[i] = cc[f];
		}
		println(m);
		for (int i = 1; i <= n; i++)
			print(gg[i] + " ");
		println();
		for (int j = 1; j <= m; j++)
			print(hh[j] + " ");
		println();
	}
}
