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<bits/stdc++.h>
using namespace std;

struct P {
  int x, y;
  bool p;
  P(){}
  P(int x, int y, bool p) : x(x), y(y), p(p){}
};

int main()
{
  P red[101];
  P blue[101];
  int n, cnt = 0;
  cin>> n;
  
  for(int i = 0; i < n; i++) {
    int x, y;
    cin>> x >> y;
    red[i] = P(x, y, false);
  }
  for(int i = 0; i < n; i++) {
    int x, y;
    cin>> x >> y;
    blue[i] = P(x, y, false);
  }
  
  for(int i = 0; i < n; i++) {
    int mn = 1145141919;
    int idxb = 0, idxr = 0;
    for(int j = 0; j < n; j++) {
      if(blue[j].p) continue;
      P tar = blue[j];
      for(int k = 0; k < n; k++) {
        if(red[k].p) continue;
        if(tar.x > red[k].x && tar.y > red[k].y) {
          if(mn > tar.x - red[k].x + tar.y - red[k].y) {
            mn = tar.x + tar.y - red[k].x - red[k].y;
            idxb = j;
            idxr = k;
          }
        }
      }
    }
    if(mn == 1145141919) continue;
    blue[idxb].p = true;
    red[idxr].p = true;
    cnt++;
  }
  
  cout<< cnt <<endl;
  
  return 0;
}