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: Runtime Error)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
	public static class Pair{
		int i, j;
		public Pair(int i, int j) {
			this.i = i;
			this.j = j;
		}
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		int[] a = new int[n];
		int[] b = new int[n];
		int[] c = new int[n];
		int[] d = new int[n];
		List<Pair> p = new ArrayList<Pair>();
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(sc.next());
			b[i] = Integer.parseInt(sc.next());
		}
		for (int i = 0; i < n; i++) {
			c[i] = Integer.parseInt(sc.next());
			d[i] = Integer.parseInt(sc.next());
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (a[i] < c[j] && b[i] < d[j])
					p.add(new Pair(i, j));
			}
		}
		Pair[] q = p.toArray(new Pair[p.size()]);
		int[] s = new int[q.length];
		int[] t = new int[q.length];
		for (int i = 0; i < q.length; i++) {
			s[i] = q[i].i;
			t[i] = q[i].j;
		}
		int max = 0;
		for (int i = 0; i < 1<<q.length; i++) {
			String o = "%" + String.valueOf(q.length) + "s";
			String u[] = String.format(o, Integer.toBinaryString(i)).replaceAll(" ", "0").split("");
			int[] v = new int[q.length];
			for (int j = 0; j < q.length; j++) {
				v[j] = Integer.parseInt(u[j]);
			}
			boolean able = true;
			for (int j = 0; j < q.length; j++) {
				if (v[j] == 1) {
					for (int k = 0; k < q.length; k++) {
						if (v[k] == 1 && k != j) {
							if (s[j] == s[k] || t[j] == t[k]) {
								able = false;
								break;
							}
						}
					}
				}
				if (!able)
					break;
			}
			if (able) {
				int count = 0;
				for (int j = 0; j < q.length; j++) {
					count += v[j];
				}
				max = Math.max(max, count);
			}
		}
		System.out.println(max);
	}
}