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.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class D {

	public static void main(String[] args) throws IOException {
		Scanner in = new Scanner(System.in);
		StringBuilder sb = new StringBuilder();
		int n = in.nextInt();
		int arr[] = new int[n + 5];
		TreeSet<Integer> ts = new TreeSet<>();
//		TreeSet<Integer> existing = new TreeSet<>();
//		TreeSet<Integer>[] pos = new TreeSet[n + 5];// integer to its positions
//													// in the arr
//		for (int i = 0; i < n + 5; i++)
//			pos[i] = new TreeSet<>();
		for (int i = 1; i <= n; i++) {
			arr[i] = in.nextInt();
			ts.add(arr[i]);
//			existing.add(arr[i]);
//			pos[arr[i]].add(i + 1);
		}
		int m_ans = ts.size();
		int h[] = new int[m_ans + 1];
		int number[] = new int[n + 5];
		int g[] = new int[n];
		boolean f = true;
		int cnt = 1;
		while(!ts.isEmpty()){
			int x = ts.pollFirst();
			h[cnt] = x;	//x-> f(x)
			if(arr[x] != x)
				f = false;
			number[x] = cnt;
			cnt++;
		}
		if (!f) {
			System.out.println(-1);
		} else {
			sb.append(m_ans).append("\n");
			for (int i = 1; i <= n; i++) {
				sb.append(number[arr[i]]).append(" ");
			}
			sb.append("\n");
			for (int i = 1; i <= m_ans; i++)
				sb.append(h[i]).append(" ");
			System.out.println(sb);
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}
	}

}
