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>
#define MOD 1000000009+7

int main(){
	int N,i,j,ans=0,tmp;
	scanf("%d",&N);
	int a[N],b[N],c[N],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-1;i++){
		for(j=0;j<N-i-1;j++){
			if(a[j]>a[j+1]){
				tmp=a[j];
				a[j]=a[j+1];
				a[j+1]=tmp;
				tmp=b[j];
				b[j]=b[j+1];
				b[j+1]=tmp;
			}
			if(c[j]>c[j+1]){
				tmp=c[j];
				c[j]=c[j+1];
				c[j+1]=tmp;
				tmp=d[j];
				d[j]=d[j+1];
				d[j+1]=tmp;
			}
		}
	}
	
	for(i=0;i<N;i++){
		tmp=-1;
		for(j=0;a[j]<c[i];j++){
			if(b[j]<d[i] && b[j]>(tmp==-1 ? -1 : b[tmp])) tmp=j;
		}
		if(tmp>=0){
			ans++;
			b[tmp]=200;
		}
	}
	
	printf("%d",ans);
	
	return 0;
}