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: Wrong Answer)

#include <stdio.h>

int main(void)
{
	int i, j, n, tmp, tmp2, cont = 0;
	int a[3][200], b[3][200];

	scanf("%d", &n);

	for(i = 0; i < 2*n; i++){
		scanf("%d%d", &tmp, &tmp2);
		a[0][tmp] = b[0][tmp2] = tmp;
		a[1][tmp] = b[1][tmp2] = tmp2;
		if(i < n){
			a[2][tmp] = 1;
			b[2][tmp2] = 1;
		}
		else{
			a[2][tmp] = 2;
			b[2][tmp2] = 2;
		}
	}

	for(i = 0; i < 2*n-1; i++){
		if(a[2][i] == 1){
			for(j = 0; j < 2*n; j++){
				if(((b[2][j] == 2) && (a[2][i] == 1)) && ((a[0][i] < b[0][j]) && (a[1][i] < b[1][j]))){
					cont++;
					b[2][j] = 0;
					break;
				}
			}
		}
	}

	printf("%d\n", cont);

	return 0;
}