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>
#include<string.h>
#include<math.h>
#include<limits.h>

int max(int a, int b){
  return a >= b ? a : b;
}
int min(int a, int b){
  return b >= a ? a : b;
}
int sei(int a){
  return a>0 ? a : 0;
}
int factorial(int n) {
    if (n > 0) return n*factorial(n - 1);
		else return 1;
}
int compare_up_int(const void *a, const void *b){
   return *(int*)a - *(int*)b;
}
int compare_down_int(const void *a, const void *b){
   return *(int*)b - *(int*)a;
}
int euclid(int a, int b){
  if(a < b){int tmp = a;a = b;b = tmp;}
  int r = a % b;
  if(r < 0) r += b;
  while(r != 0){
    a = b;b = r;r = a % b;
    if(r < 0) r += b;
  }
  return b;
}

int main(){
  int N,count=0;
  scanf("%d",&N);
  int ax[N],ay[N],bx[N],by[N];
  for(int i=0;i<N;i++) scanf("%d %d",&ax[i],&ay[i]);
  for(int i=0;i<N;i++) scanf("%d %d",&bx[i],&by[i]);

  for(int i=0;i<N;i++){
    for(int j=i+1;j<N;j++){
      if(ax[i]>ax[j]){
        int tmp1 = ax[i];ax[i]=ax[i+1];ax[i+1]=tmp1;
        int tmp2 = ay[i];ay[i]=ay[i+1];ay[i+1]=tmp2;
      }
    }
  }

  for(int i=0;i<N;i++){
    for(int j=i+1;j<N;j++){
      if(bx[i]<bx[j]){
        int tmp1 = bx[i];bx[i]=bx[i+1];bx[i+1]=tmp1;
        int tmp2 = by[i];by[i]=by[i+1];by[i+1]=tmp2;
      }
    }
  }
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j--){
      if(bx[i]>ax[j] && by[i]>ay[j]){
        count++;
        ax[j]=30000;ay[j]=30000;
        break;
      }
    }
  }
  printf("%d",count);
  return 0;
}