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.Scanner;

public class Main {
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();
		int map[][] = new int[N * 2][N * 2];
		int tmpx = 0;
		int tmpy = 0;
		for (int i = 0; i < N; i++) {
			tmpx = sc.nextInt();
			tmpy = sc.nextInt();
			map[tmpx][tmpy] = -1;
		}
		for (int i = 0; i < N; i++) {
			tmpx = sc.nextInt();
			tmpy = sc.nextInt();
			map[tmpx][tmpy] = 1;
		}
		int ans = 0;
		int tmp = 0;
		int diff = 0;
		int tmpk = 0;
		int tmpl = 0;

		for (int i = 0; i < 2 * N; i++) {
			for (int j = 0; j < 2 * N; j++) {
				if (map[i][j] != 1) {
					continue;
				}
				diff = 40000;
				tmp = 0;
				for (int k = i - 1; k > -1; k--) {
					for (int l = j - 1; l > -1; l--) {
						if (map[k][l] == -1) {
							if (diff > (i - k) + (j - l)) {
								diff = (i - k) + (j - l);
								tmpk = k;
								tmpl = l;
								tmp = 1;
							}
						}
					}
				}
				if (tmp == 1) {
					ans += 1;
					map[tmpk][tmpl] = 0;
				}
			}
		}

		System.out.println(ans);
	}
}