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.*;
import java.util.*;
import java.math.*;
import java.text.*;
public class Main
{
@SuppressWarnings("unchecked")
public static void main(String[] args) throws IOException 
{
//BufferedReader in = new BufferedReader(new FileReader(".\\mmk\\input.txt"));
//PrintWriter out=new PrintWriter("output.txt");
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

int n,x,i,cur_pi,total_pis,m;
boolean found;


n=Integer.parseInt(in.readLine().trim());

int[] f=new int[n+1];
int[] g=new int[n+1];
int[] h=new int[1000001];

String str=in.readLine().trim();
String[] token=str.split(" ");

for(i=1;i<=n;i++)
{
    f[i]=Integer.parseInt(token[i-1]);
}


cur_pi=1;
total_pis=0;
found=true;


for(x=1;x<=n;x++)
{
    if(g[x]!=0)
    {
        cur_pi=g[x];
    }
    else if(g[f[x]]!=0)
    {
        cur_pi=g[f[x]];
    }
    else
    {
        cur_pi=++total_pis;
        h[cur_pi]=f[x];
    }


    if(g[x]!=0 && g[x]!=cur_pi)
    {
        found=false;
        break;
    }
    if(g[f[x]]!=0 && g[f[x]]!=cur_pi)
    {
        found=false;
        break;
    }
    if(h[cur_pi]!=f[x])
    {
        found=false;
        break;
    }


    g[x]=cur_pi;
    g[f[x]]=cur_pi;
}



if(found)
{

    m=0;

    for(i=1;i<=1000000;i++)
    {
        if(h[i]!=0)
        {
            m++;
        }
        else
        {
            break;
        }
    }


    out.println(m);
    out.print(g[1]);

    for(i=2;i<=n;i++)
    {
        out.print(" "+g[i]);
    }

    out.println();

    out.print(h[1]);


    for(i=2;i<=m;i++)
    {
        out.print(" "+h[i]);
    }

    out.println();
}
else
{
    out.println(-1);
}



out.flush();
out.close();
}
}