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.io.PrintWriter;
import java.util.HashSet;
import java.util.StringTokenizer;

public class D {

	public static void main(String[] args) throws Exception {
		FastScanner scan = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = scan.nextInt();
		int[] f = scan.nextIntArray(n);
		HashSet<Integer> u = new HashSet<>();
		boolean good = true;
		for(int i = 0; i < n; i++) {
			u.add(f[i]);
			if(f[f[i]-1] != f[i]) good = false;
		}
		if(!good){
			out.println(-1);
			out.close();
			return;
		}
		int m = u.size();
		int[] g = new int[n];
		int[] h = new int[m];
		int c = 0;
		for(int i : u) h[c++] = i;
		int[] o = new int[n+1];
		for(int i = 0; i < m; i++) o[h[i]] = i+1;
		for(int i = 0; i < n; i++) g[i] = o[f[i]];
		out.println(m);
		for(int i = 0; i < n; i++) out.print(g[i]+" ");
		out.println();
		for(int i = 0; i < m; i++) out.print(h[i]+" ");
		out.println();
		out.close();
	}

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e){e.printStackTrace();}
		}

		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}

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

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

		public double nextDouble() {return Double.parseDouble(next());}

		public String nextLine() {
			String line = "";
			try {line = br.readLine();} 
			catch (Exception e) {e.printStackTrace();}
			return line;
		}

		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}

		public long[] nextLongArray(int n){
			long[] a = new long[n];
			for(int i = 0; i < n; i++) a[i] = nextLong();
			return a;
		}

		public double[] nextDoubleArray(int n){
			double[] a = new double[n];
			for(int i = 0; i < n; i++) a[i] = nextDouble();
			return a;
		}

		public char[][] nextGrid(int n, int m){
			char[][] grid = new char[n][m];
			for(int i = 0; i < n; i++) grid[i] = next().toCharArray();
			return grid;
		}
	}

}