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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main()
{
	int i, N, a[101], b[101], c[101], d[101];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d %d", &(a[i]), &(b[i]));
	for (i = 1; i <= N; i++) scanf("%d %d", &(c[i]), &(d[i]));
	
	int j, adj[101][101] = {};
	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++) {
			if (a[i] < c[j] && b[i] < d[j]) adj[i][j] = 1;
		}
	}
	
	int m, u, v, w, M[201] = {}, par[201], q[201], head, tail;
	for (m = 0; m <= N; m++) {
		for (i = 1; i <= N * 2; i++) par[i] = 0;
		for (i = 1; i <= N; i++) {
			if (par[i] != 0 || M[i] != 0) continue;
			par[i] = i;
			q[0] = i;
			for (head = 0, tail = 1; head < tail; head++) {
				u = q[head];
				if (u <= N) {
					for (w = N + 1; w <= N * 2; w++) {
						if (adj[u][w-N] == 1) {
							if (M[w] == 0) {
								for (v = w, par[w] = u; par[v] != v; v = par[v]) {
									if (v > N) {
										M[v] = par[v];
										M[par[v]] = v;
									}
								}
								break;
							} else if (par[w] == 0) {
								par[w] = u;
								q[tail++] = w;
							}
						}
					}
					if (w <= N * 2) break;
				} else {
					w = M[u];
					if (par[w] == 0) {
						par[w] = u;
						q[tail++] = w;
					}
				}
			}
			if (head < tail) break;
		}
		if (i > N) break;
	}
	
	printf("%d\n", m);
	fflush(stdout);
	return 0;
}