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: Time Limit Exceeded)

import java.util.*;

public class Main {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		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 k=0;k<N;k++) {
			A[k] = sc.nextInt();
			B[k] = sc.nextInt();
		}
		for(int k=0;k<N;k++) {
			C[k] = sc.nextInt();
			D[k] = sc.nextInt();
		}
		// 出力
		System.out.println(solve(A,B,C,D));
		sc.close();
	}
	
	public static String solve(int[] A, int[] B, int[] C, int[] D) {
		int N = A.length;
		List<List<Integer>> ok_blues = new ArrayList<>();
		PriorityQueue<List<Integer>> q = new PriorityQueue<>((a,b)-> a.size()-b.size());
		for(int k=0;k<N;k++) {
			List<Integer> okps = new ArrayList<>();
			for(int l=0;l<N;l++) {
				if(A[k] < C[l] && B[k] < D[l]) {
					okps.add(l);
				}
			}
			ok_blues.add(okps);
			q.add(okps);
		}
		//System.out.println(ok_blues);
		int ret = use_blue(q);
		return String.format("%d", ret);
	}
	private static int use_blue(PriorityQueue<List<Integer>> q) {
//		System.out.println("queue:" + q);
		List<Integer> p = q.poll();
		if(p == null) {
			return 0;
		}
		while(p.isEmpty() && !q.isEmpty()) {
			p = q.poll();
		}
		int max=0;
		int cnt = use_blue(new PriorityQueue<>(q));
		if(cnt > max) {
			max = cnt;
		}
//		System.out.println("---");
		for(Integer k : p) {
			PriorityQueue<List<Integer>> next = new PriorityQueue<>((a,b)-> a.size()-b.size());
			q.forEach((list) -> {
				List<Integer> newL = new ArrayList<Integer>(list);
				newL.remove(new Integer(k));
				next.add(newL);
			});
//			System.out.println("next:" + next);
			int tgt = use_blue(next)+1;
			if(tgt > max) {
				max = tgt;
			}
		}
		return max;
	}

}