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: Accepted)

/*
    - N red points + N blue points
    - Sort all points by x
    - Then, for each red points, find the blue point such that
    - xBlue > xRed, yBlue > yRed, and yBlue as close as possible
*/
#include <bits/stdc++.h>
using namespace std;
 
struct Point {
  int x, y;
};
 
int main() {
	
  // Read input
  int n, count = 0; 
  cin >> n;
  vector<Point> r(n);
  vector<Point> b(n);
  vector<int> bvisited(n, 0);
  for (int i = 0; i < n; ++i) cin >> r[i].x >> r[i].y;
  for (int i = 0; i < n; ++i) cin >> b[i].x >> b[i].y; 
  
  // Solve
  sort(r.begin(), r.end(), [] (const Point &a, const Point &b) {
    return (a.x < b.x) || (a.x == b.x && a.y < b.y);
  });
  sort(b.begin(), b.end(), [] (const Point &a, const Point &b) {
    return (a.x < b.x) || (a.x == b.x && a.y < b.y);
  }); 
  reverse(r.begin(), r.end());
  reverse(b.begin(), b.end());
 
  for (int i = 0; i < n; ++i) {
    int min_y = 2*n;
    int min_j = -1;
    for (int j = 0; j < n; ++j) {
      if (!bvisited[j])  {
        bool check_x = (r[i].x < b[j].x);
        bool check_y = (r[i].y < b[j].y);
        if (check_x && check_y) {
          if (b[j].y < min_y) {
            min_y = b[j].y;
            min_j = j;
          } 
        }
      } 
    }
 
    //cout << r[i].x << ' ' << r[i].y << "\n";
    //cout << i << ' ' << min_j << "\n\n";
    
    if (min_j != -1 && bvisited[min_j] == 0) {
      bvisited[min_j] = 1;
      count++;
    }
  }
 
  cout << count << endl;
 
	return 0;
}