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 <iostream>
#include <map>
#include <string>

using namespace std;
int main(int argc, char *argv[])
{
	int n;
	map<int, int> red;
	map<int, int> blue;

	cin >> n;
	for (int i = 0; i < n; i++){
		int a, b;
		cin >> a >> b;
		red[a] = b;
	}
	for (int i = 0; i < n; i++){
		int c, d;
		cin >> c >> d;
		blue[c] = d;
	}




	int ans1 = 0;
	map<int, int>::reverse_iterator redIte, blueIte;
	redIte = red.rbegin();
	blueIte = blue.rbegin();
	while(redIte != red.rend()){
		while(blueIte != blue.rend()){
			if (redIte->first < blueIte->first){
				ans1++;
				blueIte++;
				break;
			}
			else{
				blueIte++;
			}
		}
		redIte++;
	}

	int ans2 = 0;
	redIte = red.rbegin();
	blueIte = blue.rbegin();
	while (blueIte != blue.rend()){
		while (redIte != red.rend()){
			if (redIte->first < blueIte->first){
				ans2++;
				redIte++;
				break;
			}
			else{
				redIte++;
			}
		}
		blueIte++;
	}

	if (ans1 > ans2){
		cout << ans1 << endl;
	}
	else{
		cout << ans2 << endl;
	}
	return 0;
}