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.*;
import java.util.*;

public class ProblemD {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastScanner input = new FastScanner();
		int n = input.nextInt();
		int[] fX = new int[n];
		for (int a = 0; a < n; a++) {
			fX[a] = input.nextInt();
		}
		int numberInPlace = 0;
		TreeSet<Integer> set = new TreeSet<Integer>();
		for (int a = 0; a < n; a++) {
			set.add(fX[a]);
			if (fX[a] == a + 1) {
				numberInPlace++;
			}
		}
		if (numberInPlace == set.size()) {
			// possible
			int m = numberInPlace;
			int[] hResults = new int[m];
			int[] gResults = new int[n];
			int count = 0;
			for (Integer i : set) {
				hResults[count] = i;
				count++;
			}
			for(int a= 0; a < m; a++){
				gResults[hResults[a]-1] = a+1;
			}
			for(int a = 0; a < n; a++){
				if(gResults[a] == 0){
					int resultNeeded = fX[a];
					int idx = Arrays.binarySearch(hResults, resultNeeded);
					gResults[a] = idx+1;
				}
			}
			System.out.println(m);
			System.out.println(print(gResults));
			System.out.println(print(hResults));
		} else {
			// impossible
			System.out.println(-1);
		}
	}

	private static String print(int[] gResults) {
		StringBuilder output = new StringBuilder();
		for(int a = 0; a < gResults.length; a++){
			output.append(gResults[a] + " ");
		}
		return output.toString();
	}

	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(Reader in) {
			br = new BufferedReader(in);
		}

		public FastScanner() {
			this(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

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

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

		String readNextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int[] readIntArray(int n) {
			int[] a = new int[n];
			for (int idx = 0; idx < n; idx++) {
				a[idx] = nextInt();
			}
			return a;
		}

		long[] readLongArray(int n) {
			long[] a = new long[n];
			for (int idx = 0; idx < n; idx++) {
				a[idx] = nextLong();
			}
			return a;
		}
	}
}