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>
struct point
{
    int a,b;
};
int cmp1(const void *p1,const void *p2)
{
    struct point *c=(struct point *)p1;
    struct point *d=(struct point *)p2;
    if(c->a!=d->a)
        return (c->a - d->a);
    else
        return (c->b -d->b);
}
int cmp2(const void *p1,const void *p2)
{
    struct point *c=(struct point *)p1;
    struct point *d=(struct point *)p2;
    if(c->b!=d->b)
        return (c->b - d->b);
    else
        return (c->a -d->a);
}
int main()
{
    struct point e[1000],f[1000],u[1000];
    int i,j,h,g,l,n;
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        scanf("%d %d",&e[i].a,&e[i].b);
    }
    for(i=0; i<n; i++)
    {
        scanf("%d %d",&f[i].a,&f[i].b);
        u[i].a=f[i].a;
        u[i].b=f[i].b;
    }
    qsort(e,n,sizeof(e[0]),cmp1);
    qsort(f,n,sizeof(f[0]),cmp1);
    for(i=0,h=0; i<n; i++)
    {
        for(j=0; j<n; j++)
        {
            if(e[i].a<f[j].a&&e[i].b<f[j].b)
            {
                h++;
                //printf("%d %d %d %d\n",e[i].a,e[i].b,f[j].a,f[j].b);
                //i++;
                f[j].a=0;
                f[j].b=0;
               break;
            }
        }
    }
    //printf("%d\n",h);
    qsort(e,n,sizeof(e[0]),cmp2);
    qsort(u,n,sizeof(u[0]),cmp2);
    /*for(i=0;i<n;i++)
    {
        printf("%d %d %d %d\n",e[i].a,e[i].b,f[i].a,f[i].b);
    }*/
    for(i=0,g=0; i<n; i++)
    {
        for(j=0; j<n; j++)
        {
            if(e[i].a<u[j].a&&e[i].b<u[j].b)
            {
                g++;
                //printf("%d %d %d %d\n",e[i].a,e[i].b,f[j].a,f[j].b);
                //i++;
                u[j].a=0;
                u[j].b=0;
               break;
            }
        }
    }
    //printf("%d\n",g);
    l= h > g ? h : g;
    printf("%d\n",l);
    return 0;
}