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 <sstream>
#include<iostream>
#include<string>
#include<algorithm> 
using namespace std;
int main(){
  int n,p[205][205]={0},x,y,i,j,k,l,sum=0,my=0,mx=0;
  cin>>n;
  for(i=1;i<=n;i++){
  	cin>>y>>x;
  	p[y][x]=1;
  }
  for(i=1;i<=n;i++){
  	cin>>y>>x;
  	p[y][x]=2;
  	my=max(my,y);
  	mx=max(mx,x);
   }
   for(i=0;i<=my;i++){
  	 for(j=0;j<=mx;j++){
  		if(p[i][j])
		  cout<<p[i][j];
		else
		  cout<<'.';
	  }
	  cout<<endl;
    }
  for(i=my-1;i>=0;i--){
  	for(j=mx-1;j>=0;j--){
  		if(p[i][j]==1){
  			for(k=j+1;k<=mx;k++){
  				for(l=i+1;l<=my;l++){
  					if(p[l][k]==2){
  						p[l][k]=0;
  						p[i][j]=0;
  						sum++;
  						break;
					  }
				  }
				if(l<=mx)
				  break;
			  }
		  }
	  }
  }
  cout<<sum<<endl;
  return 0;
}