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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int[] f = new int[n + 1];
		int[] g = new int[n + 1];
		int m = 0;
		TreeMap<Integer, Integer> h = new TreeMap<Integer, Integer>();
		Integer hIndex = 1;
		for (int i=1;i<=n;i++) {
			f[i] = Integer.parseInt(st.nextToken());
			if (f[i] == i) {
				h.put(f[i], hIndex);
				hIndex++;
			}
		}
		
		m = h.size();
		boolean poss = true;
		if (m == 0) {
			poss = false;
		} else {
			for (int i=1;i<=n;i++) {
				hIndex = h.get(f[i]);
				if (hIndex == null) {
					poss = false;
					break;
				} else {
					g[i] = hIndex;
				}
			}
		}
		
		if (poss) {
			System.out.println(m);
			for (int i=1;i<=n;i++) {
				System.out.print(g[i]);
				if (i < n) {
					System.out.print(" ");
				} else {
					System.out.println();
				}
			}

			for (int key : h.keySet()) {
				System.out.print(key + " ");
			}System.out.println();
			
		} else {
			System.out.println(-1);
		}
		
		br.close();
    }
}