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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "stdio.h"
#include "stdlib.h"
#include "iostream"
#include "math.h"
#include <string.h>

//#define "limits.h"
using namespace std;




int main()
{
	int n, i, j, k=0,co=0;
	int a[200], b[200], c[200], d[200],x[20000],fr[200],fb[200],r[20000],bb[20000];
	cin >> n;
	for (i = 1;i <= n;i++) { cin >> a[i] >> b[i]; }
	for (i = 1;i <= n;i++) { cin >> c[i] >> d[i]; }
	for (i = 1;i <= n;i++)for (j = 1;j <= n;j++) {
		if (a[i] < c[j] && b[i] < d[j]) { k++;x[k] = (c[j] - a[i]) + (d[j] - b[i]);r[k] = i;bb[k] = j; }
		//else { x[i][j] = 0; }
	}
	//printf("%d\n", x[0]);
	for (i = 1;i <= n;i++) { fr[i] = 0;fb[i] = 0; }
	for (j = 1;j <= 500;j++){
		for (i = 1;i <= k;i++) {
			if (j == x[i]&&fr[r[i]]==0&&fb[bb[i]]==0) {
				co++;
				fr[r[i]] = 1;
				fb[bb[i]] = 1;
			}
		}

	}
	printf("%d\n", co);
	return 0;
}