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.Comparator;
import java.util.Scanner;

public class Main {

	public static class Pair {
		int x, y;
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		int[] b = new int[n];
		int[] c = new int[n];
		int[] d = new int[n];
		for(int i = 0 ; i < n ; i++) {
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
		}
		for(int i = 0 ; i < n ; i++) {
			c[i] = sc.nextInt();
			d[i] = sc.nextInt();
		}
		Pair[] p1 = new Pair[n];
		Pair[] p2 = new Pair[n];
		for(int i = 0 ; i < n ; i++) {
			p1[i] = new Pair(a[i], b[i]);
			p2[i] = new Pair(c[i], d[i]);
		}
		Arrays.sort(p1, new Comparator<Pair>() {
			public int compare(Pair p11, Pair p12) {
				if(p11.x == p12.x) {
					return p11.x - p12.x;
				}
				return p11.y - p12.y;
			}
		});
		Arrays.sort(p2, new Comparator<Pair>() {
			public int compare(Pair p21, Pair p22) {
				if(p21.x == p22.x) {
					return p21.x - p22.x;
				}
				return p21.y - p22.y;
			}
		});
		long ans = 0;
//		for(int i = 0 ; i < n ; i++) {
//			System.out.println("(" + p1[i].x + ", " + p1[i].y + ")");
//		}
//		for(int i = 0 ; i < n ; i++) {
//			System.out.println("(" + p2[i].x + ", " + p2[i].y + ")");
//		}
		for(int i = 0 ; i < n ; i++) {
			for(int j = 0 ; j < n ; j++) {
				if(p1[i].x == -1 || p1[j].x == -1) continue;
				if(p1[i].x < p2[j].x && p1[i].y < p2[j].y) {
					ans++;
					p1[i] = new Pair(-1, -1);
					p2[j] = new Pair(-1, -1);
				}
			}
		}
		System.out.println(ans);
	}
}