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.util.*;
public class xx {
public static void main(String[] args){
	Scanner sc=new Scanner(System.in);
int a=sc.nextInt();
int f[]=new int[a];
int g[]=new int[a];
int h[]=new int[a];
int l=0;
int m=0;
HashMap<Integer,Integer> h1=new HashMap<Integer,Integer>();
for(int i=1;i<=a;i++){
f[i-1]=sc.nextInt();
if(f[i-1]==i){
	h[l]=i;
	g[i-1]=l+1;
  h1.put(i, l);
	l++;
	m++;
}
}
if(m==0) {System.out.println(-1); return;}
for(int i=0;i<a;i++){
	if(g[i]==0){
		if(!h1.containsKey(f[i])){
			System.out.println("-1"); return;
		}
	g[i]=h1.get(f[i])+1;}
	
}
System.out.println(m);
for(int i=0;i<a;i++){
	System.out.print(g[i]+" ");
	
}System.out.println();
for(int i=0;i<m;i++){
	System.out.print(h[i]+" ");
}

}}