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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        Point[] r = new Point[N];
        for (int i = 0; i < N; i++) {
            Point p = new Point();
            p.x = scan.nextInt();
            p.y = scan.nextInt();
            r[i] = p;
        }
        Point[] b = new Point[N];
        for (int i = 0; i < N; i++) {
            Point p = new Point();
            p.x = scan.nextInt();
            p.y = scan.nextInt();
            b[i] = p;
        }
        Arrays.sort(r, new PointComparator());
        Arrays.sort(b, new ReversePointComparator());
        int r_l = 0;
        int b_l = 0;
        PointComparator c = new PointComparator();
        int answer = 0;
        boolean[] d = new boolean[N];
        Arrays.fill(d, false);
        while (r_l < N) {
            Point p1 = r[r_l];
            b_l = 0;
            while (b_l < N) {
                if (d[b_l]) {
                    b_l += 1;
                    continue;
                }
                Point p2 = b[b_l];
                if (c.compare(p1, p2) < 0) {
                    answer += 1;
                    d[b_l] = true;
                    break;
                }
                b_l += 1;
            }
            r_l += 1;
        }
        System.out.println(answer);
    }
    class Point {
        int x;
        int y;
    }
    class PointComparator implements Comparator<Point> {
        @Override
        public int compare(Point o1, Point o2) {
            if (o1.x == o2.x) {
                return Integer.compare(o1.y, o2.y);
            }
            return Integer.compare(o1.x, o2.x);
        }
    }
    class ReversePointComparator implements Comparator<Point> {
        @Override
        public int compare(Point o1, Point o2) {
            if (o1.x == o2.x) {
                return -Integer.compare(o1.y, o2.y);
            }
            return -Integer.compare(o1.x, o2.x);
        }
    }
}