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>

typedef struct{
  int x;
  int y;
  int color;
} Dot;


void sort(int n,Dot dot[200]){
  int i,j;
  Dot temp;
  for(i = 0;i < n;i++){
    for(j = n-1;j > i;j--){
      if(dot[j].y < dot[j-1].y || (dot[j].y == dot[j-1].y && dot[j].x < dot[j-1].x)){
        temp.x = dot[j-1].x;
        temp.y = dot[j-1].y;
        temp.color = dot[j-1].color;
        dot[j-1].x = dot[j].x;
        dot[j-1].y = dot[j].y;
        dot[j-1].color = dot[j].color;
        dot[j].x = temp.x;
        dot[j].y = temp.y;
        dot[j].color = temp.color;
      }
    }
  }
}

int main(){
  int n,i,j,k,red_list[200],red_list_p,ans;
  Dot dot[200];

  scanf("%d",&n);
  for(i = 0;i < n*2;i++){
    scanf("%d %d",&dot[i].x,&dot[i].y);
    if(i < n){
      dot[i].color = 0;
    }else{
      dot[i].color = 1;
    }
  }

  sort(n*2,dot);

  ans = 0;
  red_list_p = 0;
  ans = 0;
  for(i = 0;i < n*2;i++){
    if(dot[i].color == 0){
      red_list_p++;
      red_list[red_list_p] = i;
    }else{
      for(j = red_list_p;j > 0;j--){
        if(dot[red_list[j]].x < dot[i].x){
          ans++;
          for(k = j;k <red_list_p;k++){
            red_list[k] = red_list[k+1];
          }
          red_list_p--;
          break;
        }
      }
    }
  }

  printf("%d\n",ans);

  return(0);
}