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>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}

ll d[210];
int main(){
	int n;
	scanf("%d",&n);
	rep(i,0,2*n){
		int x,y;
		scanf("%d%d",&x,&y);
		d[i]=x*2000+y*2+(i/n);
	}
	sortup(d,2*n);
	ll ans=0;
	rep(i,0,2*n)if(d[i]%2){
		int index=-1;
		int ymax=-1;
		rep(j,0,i)if(d[j]%2==0&&d[j]%2000/2>ymax){
			ymax=d[j]%2000/2;
			index=j;
		}
		if(ymax!=-1){
			ans++;
			d[index]=-1;
		}
		d[i]=-1;
	}
	printf("%lld\n",ans);
	return 0;
}