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.io.*;
import java.util.*;

class Main {

	static class Dinic {
		private List<Edge>[] g;
		private int[] level;
		private int[] iterator;

		class Edge {
			int to;
			int cap;
			int residualId;

			public Edge(int to, int cap, int residualId) {
				this.to = to;
				this.cap = cap;
				this.residualId = residualId;
			}
		}

		public Dinic(int n) {
			g = new ArrayList[n];
			level = new int[n];
			iterator = new int[n];
			for (int i = 0; i < g.length; i++) {
				g[i] = new ArrayList<>();
			}
		}

		public void addEdge(int from, int to, int cap) {
			g[from].add(new Edge(to, cap, g[to].size()));
			g[to].add(new Edge(from, 0, g[from].size() - 1)); // residual path
		}

		public long maxFlow(int s, int t) {
			long flow = 0;
			while (true) {
				bfs(s);
				if (level[t] < 0) break;
				Arrays.fill(iterator, 0);
				while (true) {
					long f = dfs(s, t, Long.MAX_VALUE);
					if (f == 0) break;
					flow += f;
				}
			}
			return flow;
		}

		private void bfs(int s) {
			Arrays.fill(level, -1);
			Queue<Integer> q = new LinkedList<>();
			level[s] = 0;
			q.offer(s);
			while (!q.isEmpty()) {
				int v = q.poll();
				for (int i = 0; i < g[v].size(); i++) {
					Edge e = g[v].get(i);
					if (e.cap > 0 && level[e.to] < 0) {
						level[e.to] = level[v] + 1;
						q.offer(e.to);
					}

				}
			}
		}

		private long dfs(int s, int t, long f) {
			if (s == t) return f;
			for (int i = iterator[s]; i < g[s].size(); i = ++iterator[s]) {
				Edge e = g[s].get(i);
				if (e.cap > 0 && level[s] < level[e.to]) {
					long d = dfs(e.to, t, Math.min(e.cap, f));
					if (d > 0) {
						e.cap -= d;
						g[e.to].get(e.residualId).cap += d;
						return d;
					}
				}
			}
			return 0;
		}
	}

	public static void main(String[] args) throws Exception {
		final Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		Dinic dinic = new Dinic(n * 2 + 2);
		for (int i = 0; i < n; i++) {
			dinic.addEdge(2 * n, i, 1);
			dinic.addEdge(i + n, 2 * n + 1, 1);
		}

		int[][] a = new int[n][2];
		int[][] b = new int[n][2];
		for (int i = 0; i < n; i++) {
			a[i][0] = sc.nextInt();
			a[i][1] = sc.nextInt();
		}
		for (int i = 0; i < n; i++) {
			b[i][0] = sc.nextInt();
			b[i][1] = sc.nextInt();
		}

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (a[i][0] < b[j][0] && a[i][1] < b[j][1]) {
					dinic.addEdge(i, n + j, 1);
				}
			}
		}

		System.out.println(dinic.maxFlow(2 * n, 2 * n + 1));
	}
}