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: Accepted)

import java.util.*;
import java.util.function.*;
import java.math.*;
public class Main {
    public static void main(String[] args){
		Scanner s = new Scanner(System.in);
		int N = s.nextInt();
		int[][] a = new int[N][3];
		int[][] b = new int[N][3];
		for(int i = 0; i < N; i++) {
			a[i][0] = s.nextInt();
			a[i][1] = s.nextInt();
			a[i][2] = 1; 
		}
		for(int i = 0; i < N; i++) {
			b[i][0] = s.nextInt();
			b[i][1] = s.nextInt();
			b[i][2] = 1;
		}
		
		Arrays.sort(b, (x, y) -> Integer.compare(x[0], y[0]));
		Arrays.sort(a, (x, y) -> Integer.compare(y[1], x[1]));
		
		// a,b : 0=x, 1=y, 2=使用済みフラグ, 3=内包する点の数
		int ret = 0;
		for(int i = 0; i < N; i++) {
			for(int j = 0; j < N; j++) {
				if(b[i][0] > a[j][0] && b[i][1] > a[j][1] && a[j][2] > 0) {
					b[i][2]--;
					a[j][2]--;
					ret++;
					break;
				}
			}
		}
		System.out.println(ret);
    }
}