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.awt.Point;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Collections;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.List;
import java.util.PriorityQueue;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(br.readLine());
        Comparator<Point> comp = (p1, p2) -> p1.x - p2.x;
        List<Point> red = new LinkedList<>();
        for(int i = 0; i < n; i++) {
            String[] in = br.readLine().split(" ");
            red.add(new Point(Integer.parseInt(in[0]), Integer.parseInt(in[1])));
        }
        Collections.sort(red, comp);
        PriorityQueue<Point> blue = new PriorityQueue<>(comp);
        for(int i = 0; i < n; i++) {
            String[] in = br.readLine().split(" ");
            blue.add(new Point(Integer.parseInt(in[0]), Integer.parseInt(in[1])));
        }

        int ans = 0;
        while(!blue.isEmpty()) {
            Point b = blue.poll();
            
            int index = -1;
            int maxy = -1;
            int i = 0;
            Point tmp;
            while(i < red.size() && (tmp = red.get(i)).x < b.x) {
                if(tmp.y < b.y && maxy < tmp.y) {
                    index = i;
                    maxy = tmp.y;
                }
                i++;
            }

            if(index != -1) {
                red.remove(index);
                ans++;
            }
        }

        System.out.println(ans);

        br.close();
    }
}