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.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();

    Point[] x = new Point[N * 2];
    Point[] y = new Point[N * 2];

    for (int j = 0; j < 2; j++) {
      for (int i = 0; i < N; i++) {
        x[i + N * j] = new Point(scanner.nextInt(), j);
        y[i + N * j] = new Point(scanner.nextInt(), j);
      }
    }

    Arrays.sort(x);
    Arrays.sort(y);

    System.out.println(Math.min(count(x, N), count(y, N)));
  }

  private static int count(Point[] a, int N) {
    int count = 0;
    for(Point p : a) {
      if (p.type == 0) {
        count++;
      } else if (count > 0) {
        count--;
      }
    }
    return N - count;
  }

  private static class Point implements Comparable<Point> {
    final int pos;
    final int type;

    private Point(int pos, int type) {
      this.pos = pos;
      this.type = type;
    }

    @Override public int compareTo(Point o) {
      return pos - o.pos;
    }
  }
}