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 <stdlib.h>
#include <string.h>
typedef struct
{
    int a;
    int b;
} yayaya;
int main()
{
    yayaya ya[101],hei[101];
    int x,y,i,j,s,temp;
    scanf("%d",&x);
    for(i=0; i<x; i++)
    {
        scanf("%d%d",&ya[i].a,&ya[i].b);
    }
    y=x;
    for(i=0; i<y; i++)
    {
        scanf("%d%d",&hei[i].a,&hei[i].b);
    }
    for(i=0; i<x; i++)
    {
        for(j=i+1; j<x; j++)
        {
            if(ya[i].a>ya[j].a)
            {
                temp=ya[i].a;
                ya[i].a=ya[j].a;
                ya[j].a=temp;
                temp=ya[i].b;
                ya[i].b=ya[j].b;
                ya[j].b=temp;
            }
        }
    }
    for(i=0; i<y; i++)
    {
        for(j=i+1; j<y; j++)
        {
            if(hei[i].a>hei[j].a)
            {
                temp=hei[i].a;
                hei[i].a=hei[j].a;
                hei[j].a=temp;
                temp=hei[i].b;
                hei[i].b=hei[j].b;
                hei[j].b=temp;
            }
        }
    }
    s=0;
    int f,min,k,h;
    for(i=0; i<x; i++)//第一步确定那个x
    {
        for(j=0; j<y; j++)//开始循环找第一位大的
        {
            if(ya[i].a<hei[j].a)//找到
            {
                for(k=j; k<y; k++)
                {
                    if(ya[i].b<hei[k].b)
                    {
                        min=hei[k].b;f=k;
                        for(h=k;h<y;h++)
                        {
                            if(min>hei[h].b&&hei[h].b>ya[i].b)
                            {
                                min=hei[h].b;
                                f=h;
                            }
                        }
                    hei[f].b=hei[f].a=-1000000;
                    s++;
                    break;
                    }
                }

                break;
            }
        }
    }
    printf("%d",s);
    return 0;
}