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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

typedef unsigned int uint;

void run(void){
  uint n;
  scanf("%u",&n);
  uint *a=(uint *)calloc(n,sizeof(uint));
  uint *b=(uint *)calloc(n,sizeof(uint));
  uint i;
  for(i=0;i<n;i++) scanf("%u",a+i);
  for(i=0;i<n;i++) scanf("%u",b+i);
  const uint block=1<<12;
  uint ans=0;
  for(i=0;i<n;i+=block){
    uint sup=i+block<=n?i+block:n;
    uint x,y,z,w;
    x=y=z=w=0;
    uint j;
    for(j=0;j+3<n;j+=4){
      for(uint k=i;k<sup;k++){
	x^=a[k]+b[j];
	y^=a[k]+b[j+1];
	z^=a[k]+b[j+2];
	w^=a[k]+b[j+3];
      }
    }
    ans^=x^y^z^w;
    for(;j<n;j++){
      for(uint k=i;k<sup;k++){
	ans^=a[k]+b[j];
      }
    }
  }
  printf("%u\n",ans);
}

int main(void){
  run();
  return 0;
}