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.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

import static java.lang.Integer.parseInt;

public class Main {

  final int MIN = -1_000_000_000;
  public static void main(String[] args) {
    new Main().run();
  }

  public void run() {

    try(BufferedReader in = new BufferedReader(new InputStreamReader(System.in))) {

      int n = parseInt(in.readLine());

      pxPoint[] reds  = new pxPoint[n];
      pxPoint[] blues = new pxPoint[n];
      String[] lines;
      for(int i=0, x=0, y=0; i<n; i++) {
        lines = in.readLine().split(" ");
        x = parseInt(lines[0]);
        y = parseInt(lines[1]);
        reds[i] = new pxPoint(x, y); 
      }
      for(int i=0, x=0, y=0; i<n; i++) {
        lines = in.readLine().split(" ");
        x = parseInt(lines[0]);
        y = parseInt(lines[1]);
        blues[i] = new pxPoint(x, y); 
      }

      Arrays.parallelSort(reds);
      Arrays.parallelSort(blues);

      boolean[] isDone = new boolean[n];

      int ans = 0;
      pxPoint p = new pxPoint(MIN,MIN);
      int maxJ = MIN;
      for(int i=0; i<n; i++) {
        for(int j=0; j<n; j++) {
          if(!isDone[j] && reds[j].x < blues[i].x) {
            if(reds[j].y >= p.y) {
              p.x = reds[j].x;
              p.y = reds[j].y;
              maxJ = j;
            }
          }
        }
        if(maxJ != MIN) {
          ans++;
          isDone[maxJ] = true;
          p.x = MIN; p.y = MIN; maxJ = MIN;
        }
      }

      System.out.println(ans);
      // for (pxPoint p : reds) {
      //   System.out.println(p.x + " " + p.y);
      // }
    }
    catch(IOException e) {
      System.err.println(e);
    }
  }

  class pxPoint extends Point implements Comparable<pxPoint> {

    public pxPoint() {
      super();
    }
    public pxPoint(int x, int y) {
      super(x, y);
    }

    // @Override
    public int compareTo(pxPoint obj) {
      if(this.x == obj.x) return 0;
      return this.x > obj.x ? 1 : -1;
    }
  }
}