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.*;

public class Main {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		Map<Integer, Integer> red_x = new TreeMap<Integer, Integer>();
		Map<Integer, Integer> red_y = new TreeMap<Integer, Integer>();
		Map<Integer, Integer> blue_x = new TreeMap<Integer, Integer>();
		Map<Integer, Integer> blue_y = new TreeMap<Integer, Integer>();
		
		for(int i = 0; i < N; i++) {
			int temp_x = sc.nextInt();
			int temp_y = sc.nextInt();
			red_x.put(temp_x, temp_y);
			red_y.put(temp_y, temp_x);
		}
		
		for(int i = 0; i < N; i++) {
			int temp_x = sc.nextInt();
			int temp_y = sc.nextInt();
			blue_x.put(temp_x, temp_y);
			blue_y.put(temp_y, temp_x);
		}
		
		int cnt1 = 0;
		ArrayList<Integer> use_key1 = new ArrayList<Integer>();
		outside : for(int red_key : red_x.keySet()) {
			for(int blue_key : blue_x.keySet()) {
				if(red_key < blue_key && red_x.get(red_key) < blue_x.get(blue_key)) {
					if(!use_key1.contains(blue_key)) {
						cnt1++;
						use_key1.add(blue_key);
						continue outside;
					}
				}
			}
		}
		
		int cnt2 = 0;
		ArrayList<Integer> use_key2 = new ArrayList<Integer>();
		outside : for(int red_key : red_y.keySet()) {
			for(int blue_key : blue_y.keySet()) {
				if(red_key < blue_key && red_y.get(red_key) < blue_y.get(blue_key)) {
					if(!use_key2.contains(blue_key)) {
						cnt2++;
						use_key2.add(blue_key);
						continue outside;
					}
				}
			}
		}
		
		System.out.println(Math.max(cnt1, cnt2));
	}	
}