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 <math.h>

int main(void){
    int n,count=0;
    int ax[205],ay[205],bx[205],by[205];
    scanf("%d",&n);
    for(int i=1; i<=n; i++){
        scanf("%d%d",&ax[i],&ay[i]);
    }    
    for(int i=1; i<=n; i++){
        scanf("%d%d",&bx[i],&by[i]);
    }

    //blueのyソート
    for(int i=1; i<=n; i++){
        int temx=bx[i],temy=by[i],temp=0;
        for(int j=i+1; j<=n; j++){
            if(temy > by[j]){
                temx=bx[j],temy=by[j],temp=j;
            }
        }
        if(temp){
            bx[temp]=bx[i],by[temp]=by[i];
            bx[i]=temx,by[i]=temy;
        }
    }

    for(int i=1; i<=n; i++){
        int temx=-1, temy=-1, temp=0;
        //blueのi※
        for(int j=1; j<=n; j++){
            //redのチェック
            if(ax[j]<bx[i] && ay[j]<by[i]){
                if(temx < ax[j]){
                    temx=ax[j],temy=ay[j],temp=j;
                }
            }
        }
        if(temp){
            count++;
            ax[temp]=300,ay[temp]=300;
        }
    }
    printf("%d\n",count);
    return 0;
}