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[][] g = new int[N*2][N*2]; // 1:red. 2:blue.
        for(int i = 0; i < N; i++){
            int x = sc.nextInt();
            int y = sc.nextInt();
            g[x][y] = 1;
        }
        for(int i = 0; i < N; i++){
            int x = sc.nextInt();
            int y = sc.nextInt();
            g[x][y] = 2;
        }
        int[][] rest = new int[N*2][N*2];
        int ans = 0;
        for(int x = 0; x < N*2; x++){
            for(int y = 0; y < N*2; y++){
                if(x > 0 && y > 0) rest[x][y] = rest[x-1][y] + rest[x][y-1] - rest[x-1][y-1];
                else if(x == 0 && y > 0) rest[x][y] = rest[x][y-1];
                else if(x > 0 && y == 0) rest[x][y] = rest[x-1][y];

                if(g[x][y] == 1){
                    rest[x][y]++;
                }else if(g[x][y] == 2 && rest[x][y] > 0){
                    ans++;
                    rest[x][y]--;
                }
            }
        }
        System.out.println(ans);
        sc.close();

    }

}