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.*;
 
public class Main{
	public static boolean[][] map = new boolean[100][100]; //[red][blue]
	public static int[] blues = new int[100];
	public static int[] reds = new int[100];
	
	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();
		}
		for(int i=0;i<N;++i){
			for(int j=0;j<N;++j){
				if(a[i]<c[j]&&b[i]<d[j]){
					map[i][j] = true;
					blues[i]++;
					reds[j]++;
				}
			}
		}
		
		int ans = 0;
		while(!noPair(N)){
			int red = biggest(N);
			int blue = -1;
			int temp = Integer.MAX_VALUE;
			for(int i=0;i<N;++i){
				if(map[red][i]&&reds[i]<temp){
					blue = i;
					temp = reds[i];
				}
			}
			
			delete(red,blue,N);
			ans++;
		}
		
		System.out.println(ans);
		return;
	}
	
	public static boolean noPair(int n){
		for(int i=0;i<n;++i){
			if(reds[i]!=0) return false;
 		}
		return true;
	}
	
	public static int biggest(int n){
		int max = Integer.MAX_VALUE;
		int ret = -1;
		for(int i=0;i<n;++i){
			if(blues[i]!=0&&blues[i]<max){
				max = blues[i];
				ret = i;
			}
		}
		return ret;
	}
	
	public static void delete(int r,int b,int n){
		for(int i=0;i<n;i++){
			if(map[r][i]){
				map[r][i] = false;
				blues[r]--;
				reds[i]--;
			}
			if(map[i][b]){
				map[i][b] = false;
				blues[i]--;
				reds[b]--;
			}
		}
	}
}