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>
int main()
{
    int n,a[105],b[105],c[105],d[105],i,j,temp,count=0,p;
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        scanf("%d%d",&a[i],&b[i]);
    }
    for(i=0; i<n; i++)
    {
        scanf("%d%d",&c[i],&d[i]);
    }
    for(i=0; i<n; i++)
    {
        for(j=i+1; j<n; j++)
        {
            if((a[j]<a[i] && a[j]<b[i]) || (b[j]<a[i] && b[j]<b[i]))
            {
                temp=a[i];
                a[i]=a[j];
                a[j]=temp;

                temp=b[i];
                b[i]=b[j];
                b[j]=temp;
            }
        }
    }

    for(i=0; i<n; i++)
    {
        for(j=i+1; j<n; j++)
        {
            if((c[j]<c[i] && c[j]<d[i]) || (d[j]<c[i] && d[j]<d[i]))
            {
                temp=c[i];
                c[i]=c[j];
                c[j]=temp;

                temp=d[i];
                d[i]=d[j];
                d[j]=temp;
            }
        }
    }

    /*for(i=0; i<n; i++)
    {
       printf("%d %d\n",a[i],b[i]);
    }
    for(i=0; i<n; i++)
    {
       printf("%d %d\n",c[i],d[i]);
    }*/
    p=-1;
    for(i=0; i<n; i++)
    {
        for(j=p+1; j<n; j++)
        {
            if(a[i]<c[j] && b[i]<d[j])
            {
                count++;
                //printf("%d\n",count);
                break;
            }
        }
        p=j;
    }
    printf("%d",count);
}