Problem:
Score : 400 points

Problem Statement

On a two-dimensional plane, there are N red points and N blue points.
The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Constraints

All input values are integers.

1 \leq N \leq 100

0 \leq a_i, b_i, c_i, d_i < 2N

a_1, a_2, ..., a_N, c_1, c_2, ..., c_N are all different.

b_1, b_2, ..., b_N, d_1, d_2, ..., d_N are all different.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
:
a_N b_N
c_1 d_1
c_2 d_2
:
c_N d_N

Output

Print the maximum number of friendly pairs.

Sample Input 1

3
2 0
3 1
1 3
4 2
0 4
5 5

Sample Output 1

2

For example, you can pair (2, 0) and (4, 2), then (3, 1) and (5, 5).

Sample Input 2

3
0 0
1 1
5 2
2 3
3 4
4 5

Sample Output 2

2

For example, you can pair (0, 0) and (2, 3), then (1, 1) and (3, 4).

Sample Input 3

2
2 2
3 3
0 0
1 1

Sample Output 3

0

It is possible that no pair can be formed.

Sample Input 4

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7

Sample Output 4

5

Sample Input 5

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9

Sample Output 5

4

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

import java.util.*;
 class Node{
static int x;
static int y;
  public Node(){
x=0;
y=0;}
}   
public class Main{
  static int[][] line;
     static int[] used;
     static int[] linker;
     static int n;
       static  Node[] a;
     static Node[] b;
     public static void main(String[] args){
         Scanner in = new Scanner(System.in);
         n=in.nextInt();
       a= new Node[n];
     b= new Node[n];
    for(int i=0;i<a.length;i++){
      a[i]=new Node();
     b[i]=new Node();               
}   
   line=new int[n][n];
   used=new int[n];
   linker=new int[n];
 for(int i = 0; i<n;i++){ a[i].x=in.nextInt(); a[i].y=in.nextInt();}
         for(int i = 0; i<n;i++){ b[i].x=in.nextInt(); b[i].y=in.nextInt();}     
         for(int i=0;i<n;i++)
          for(int j=0;j<n;j++){
             if(a[i].x<=b[j].x && a[i].y<=b[j].y)
               line[i][j]=1;
        }
       System.out.println(hungary());
     }
   public static int hungary(){
   int sum=0;
  for(int i=0; i<n;i++){
     for(int j=0;j<n;j++)
         used[j]=0;
      if(find(i)) sum++;
}
return sum;
}
 private static boolean find(int x){
for(int i=0;i<n;i++){
if(line[x][i]==1 && used[i]==0){
used[i]=1;
if(linker[i]==0 || find(linker[i])){
linker[i]=x;
return true;
}
}
}
return false;}

    }