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>
int main(void){
   int N,i,j;
   scanf("%d",&N);
   int a[N],b[N],c[N],d[N],max_x,max_y,max_n;
   
   for(i=0;i<N;i++){
       scanf("%d %d",&a[i],&b[i]);
   }
   for(i=0;i<N;i++){
       scanf("%d %d",&c[i],&d[i]);
   }
   
   //ｘの小さい順にソート
   for(i=0;i<N;i++){
       max_x = 0;
       max_y = 0;
       max_n = i;
       for(j=i;j<N;j++){
           if(max_x < a[j]){
               max_x = a[j];
               max_y = b[j];
               max_n = j;
           }
       }
       a[max_n] = a[i];
       b[max_n] = b[i];
       a[i] = max_x;
       b[i] = max_y;
   }
   
   
   for(i=0;i<N;i++){
       max_x = 0;
       max_y = 0;
       max_n = i;
       for(j=i;j<N;j++){
           if(max_x < c[j]){
               max_x = c[j];
               max_y = d[j];
               max_n = j;
           }
       }
       c[max_n] = c[i];
       d[max_n] = d[i];
       c[i] = max_x;
       d[i] = max_y;
   }
   
   //更にｙの小さい順にソート
   
   
   for(i=0;i<N;i++){
       max_x = 0;
       max_y = 0;
       max_n = i;
       for(j=i;j<N;j++){
           if(max_y < b[j]){
               max_x = a[j];
               max_y = b[j];
               max_n = j;
           }
       }
       a[max_n] = a[i];
       b[max_n] = b[i];
       a[i] = max_x;
       b[i] = max_y;
   }
   
   for(i=0;i<N;i++){
       max_x = 0;
       max_y = 0;
       max_n = i;
       for(j=i;j<N;j++){
           if(max_x < d[j]){
               max_x = c[j];
               max_y = d[j];
               max_n = j;
           }
       }
       c[max_n] = c[i];
       d[max_n] = d[i];
       c[i] = max_x;
       d[i] = max_y;
   }
   
   //探索
   
   int count=0;
   
   for(i=0;i<N;i++){
       for(j=0;j<N;j++){
           if(a[i]<c[i] && b[i]<d[i]){
               count ++;
               c[i] = 0;
               d[i] = 0;
               break;
           }
       }
   }
   
   printf("%d",count);
}