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>
#include <string.h>
#include <stdlib.h>

int main() {
    int n;
    scanf("%d", &n);

    int** red = (int**)calloc(n, sizeof(int*));
    int** blue = (int**)calloc(n, sizeof(int*));

    int i, j;
    for (i = 0; i < n; i++) {
        red[i] = (int*)calloc(3, sizeof(int));
        blue[i] = (int*)calloc(2, sizeof(int));
    }

    for (i = 0; i < n; i++) {
        scanf("%d %d", &red[i][0], &red[i][1]);
    }
    for (i = 0; i < n; i++) {
        scanf("%d %d", &blue[i][0], &blue[i][1]);
    }

    int pairs = 0;

    for (i = 0; i < n; i++) {
        int pairredid = -1;
        long bestdist = -1;
        for (j = 0; j < n; j++) {
            if (blue[i][0] - red[j][0] > 0 && blue[i][1] - red[j][1] > 0 && red[j][2] == 0) {
                double dist = (blue[i][0] - red[j][0]) * (blue[i][0] - red[j][0]) + (blue[i][1] - red[j][1]) * (blue[i][1] - red[j][1]);
                if (bestdist < 0 || dist < bestdist) {
                    bestdist = dist;
                    pairredid = j;
                }
            }
        }
        if (pairredid != -1) {
            red[pairredid][2] = 1;
            pairs++;
            printf("R: (%d, %d)\n", red[pairredid][0], red[pairredid][1]);
            printf("B: (%d, %d)\n\n", blue[i][0], blue[i][1]);
        }
    }

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

}