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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#define REP(i,n) for(int i=0;i<n;i++)
#define REPP(i,n) for(int i=1;i<=n;i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define EPS (1e-9)
#define INF (1e17)
#define PI (acos(-1))
//const double PI = acos(-1);
//const double EPS = 1e-15;
//long long INF=(long long)1E17;
#define i_7 (long long)(1e9+7)
//#define i_7 998'244'353
long mod(long a){
    long long c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
using namespace std;

bool prime_(int n){
  if(n==1){
    return false;
  }else if(n==2){
    return true;
  }else{
    for(int i=2;i<=sqrt(n);i++){
      if(n%i==0){
        return false;
      }
    }
    return true;
  }
}

long long gcd_(long long a, long long b){
  if(a<b){
    swap(a,b);
  }
  if(a%b==0){
    return b;
  }else{
    return gcd_(b,a%b);
  }
}
 
long long lcm_(long long x, long long y){
  return (x/gcd_(x,y))*y;
}

int main(){
  
  int n;
  cin>>n;
  long long a[n];
  long long b[n];
  REP(i,n){
    cin>>a[i];
  }
  REP(i,n){
    cin>>b[i];
  }
  vector<pair<long long, char>> seq;
  REP(i,n){
    seq.push_back(make_pair(a[i], 'a'));
    seq.push_back(make_pair(b[i], 'b'));
  }
  sort(ALL(seq));
  
  stack<long long> as;
  stack<long long> bs;
  
  long long ans = 1;
  REP(i,2*n){
    if(seq[i].second == 'a'){
      if(bs.empty()){
        as.push(seq[i].first);
      }else{
        ans *= (long long)bs.size();
        ans = mod(ans);
        bs.pop();
      }
    }else{
      if(as.empty()){
        bs.push(seq[i].first);
      }else{
        ans *= (long long)as.size();
        ans = mod(ans);
        as.pop();
      }
    }
  }
  
  cout<<ans<<endl;
      
  return 0;
}