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;
import java.util.Iterator;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;

class Point implements Comparable {
    public Integer x;
    public Integer y;

    public Point(int x, int y) {
        this.x = x;
        this.y = y;
    }

    public void print() {
        System.out.println(x + ", " + y);
    }

    @Override
    public int compareTo(Object o) {
        Point p = (Point) o;

        if (x.equals(p.x)) {
            return y.compareTo(p.y);
        }
        else {
            return x.compareTo(p.x);
        }
    }
}

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();

        List<Point> reds = new ArrayList<>();
        List<Point> blues = new ArrayList<>();

        for (int i = 0; i < N; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            reds.add(new Point(x, y));
        }

        for (int i = 0; i < N; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            blues.add(new Point(x, y));
        }

        Collections.sort(reds, Collections.reverseOrder());
        Collections.sort(blues, Collections.reverseOrder());

        int answer = 0;

        Iterator<Point> it = reds.iterator();
        while (it.hasNext()) {
            Point red = it.next();

            for (int j = 0; j < blues.size(); j++) {
                if (isNakayoshi(red, blues.get(j))) {
                    answer++;
                    blues.remove(blues.get(j));
                    break;
                }
            }
        }

        System.out.println(answer);
    }

    public static boolean isNakayoshi(Point red, Point blue) {
        return (red.compareTo(blue) == -1);
    }
}