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.awt.*;
import java.util.*;
import java.util.List;

import static java.lang.System.*;

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

        List<Point> redPoints = new ArrayList<>();
        List<Point> bluePoints = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            redPoints.add(new Point(sc.nextInt(), sc.nextInt()));
        }
        for (int i = 0; i < N; i++) {
            bluePoints.add(new Point(sc.nextInt(), sc.nextInt()));
        }

        redPoints.sort(new PointSort());
        bluePoints.sort(new PointSort());

        int cnt = 0;
        for (Point bp : bluePoints) {

            Point rp = null;

            for (int i = 0; i < redPoints.size(); i++) {
                Point tmprp = redPoints.get(i);
                if (tmprp.x >= bp.x) {
                    break;
                }

                // 出来るだけ大きいYのポイントを取る
                if (rp == null || (rp.y < tmprp.y && bp.y > tmprp.y)){
                    rp = tmprp;
                }
            }

            if(rp != null){
                cnt++;
                redPoints.remove(rp);
            }
        }

        out.println(cnt);
    }

    public static class PointSort implements Comparator<Point> {
        @Override
        public int compare(Point o1, Point o2) {
            return Integer.compare(o1.x, o2.x);
        }
    }
}