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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>

void maopao(int *a,int *b,int n)
{
	int i,j;
	for(i=0;i<n;i++)
    {
    	for(j=i+1;j<n;j++)
    	{
    		if(a[i]<a[j])
    		{
    			int temp;
    			temp=a[i];
    			a[i]=a[j];
    			a[j]=temp;
    			temp=b[i];
    			b[i]=b[j];
    			b[j]=temp;
			}
		}
	}
}
int main()
{
    int i,j,k,n,cnt=0;
    int a[1000],b[1000],c[1000],d[1000];
    scanf("%d",&n);
    j=k=n;
    i=0;
    while(k--)
    {
    	scanf("%d %d",&a[i],&b[i]);
    	i++;
	}
//					printf("%d\n",a[2]);
    i=0;
    while(j--)
    {
    	scanf("%d %d",&c[i],&d[i]);
    	i++;
	//	printf("%d\n",c[0]);
	}
	//	printf("%d\n",c[0]);
    maopao(a,b,n);
    maopao(c,d,n);
//	for(i=0;i<n;i++)
//    {
//		printf("%d %d\n",a[i],b[i]);
//		printf("%d %d\n",c[i],d[i]);
//	}
/*
	for(i=0;i<n;i++)
	{
		for(j=i;j<n;j++)
		{
			if(c[i]>a[j])
				if(d[i]>b[j])
				{
					b[j]=1000;
					a[j]=1000;
					cnt++;
					break;
				}
		}
	}
*/
    k=0;
	for(i=c[k];i>=0;i--)
	{
		for(j=0;j<n;j++)
		{
            if(i==a[j])
            {
                if(b[k]<d[j])
                cnt++;
                break;
            }
            if(i==c[j])
            {
                 k++;
                 break;
            }

		}


	}
    printf("%d\n",cnt);
    return 0;
}