Problem:
Score : 500 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

There are N computers and N sockets in a one-dimensional world.
The coordinate of the i-th computer is a_i, and the coordinate of the i-th socket is b_i.
It is guaranteed that these 2N coordinates are pairwise distinct.

Snuke wants to connect each computer to a socket using a cable.
Each socket can be connected to only one computer.

In how many ways can he minimize the total length of the cables?
Compute the answer modulo 10^9+7.

Constraints

1 ≤ N ≤ 10^5

0 ≤ a_i, b_i ≤ 10^9

The coordinates are integers.

The coordinates are pairwise distinct.

Input

The input is given from Standard Input in the following format:

N
a_1
:
a_N
b_1
:
b_N

Output

Print the number of ways to minimize the total length of the cables, modulo 10^9+7.

Sample Input 1

2
0
10
20
30

Sample Output 1

2

There are two optimal connections: 0-20, 10-30 and 0-30, 10-20.
In both connections the total length of the cables is 40.

Sample Input 2

3
3
10
8
7
12
5

Sample Output 2

1

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        long mod = 1000000000+7;
        int n = sc.nextInt();
        int[] a = new int[n], b = new int[n];
        for(int i=0;i<n;i++) a[i]=sc.nextInt();
        for(int i=0;i<n;i++) b[i]=sc.nextInt();
        Arrays.sort(a);
        Arrays.sort(b);
        long[] fac = new long[n+1];
        fac[0]=1;
        for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
        long ans = 1;
        int left = 0;
        int[] base,target;
        while(left<n){
            if(a[left]>b[left]){
                base = b;
                target = a;
            } else{
                base = a;
                target = b;
            }
            int right = left+1;
            while(right<n&&base[right]<target[left]) right++;
            ans = ans*fac[right-left]%mod;
            left = right;
        }
        System.out.println(ans);
    }
}