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 <string>
#include <vector>
#include <algorithm>

using namespace std;

struct xyz
{
	int x;
	int y;
};

int main()
{
	int n;
	cin >> n;
	
	vector<xyz> red(n);
	vector<xyz> blue(n);
	for(int i = 0; i < n; i++)
	{
		cin >> red[i].x >> red[i].y;
	}
	for(int i = 0; i < n; i++)
	{
		cin >> blue[i].x >> blue[i].y;
	}
	
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	
	int ans1 = 0;
	vector<bool> flag1(n, false);
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			if(red[j].x < blue[i].x && red[j].y < blue[i].y && flag1[j] != true)
			{
				ans1++;
				flag1[j] = true;
				break;
			}
		}
	}
	
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	
	int ans2 = 0;
	vector<bool> flag2(n, false);
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			if(red[j].x < blue[i].x && red[j].y < blue[i].y && flag2[j] != true)
			{
				ans2++;
				flag2[j] = true;
				break;
			}
		}
	}
	
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	
	int ans3 = 0;
	vector<bool> flag3(n, false);
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			if(red[j].x < blue[i].x && red[j].y < blue[i].y && flag3[j] != true)
			{
				ans3++;
				flag3[j] = true;
				break;
			}
		}
	}
	
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	sort(red.begin(), red.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.y < b.y;});
	sort(blue.begin(), blue.end(), [](const xyz &a, const xyz &b){return a.x < b.x;});
	
	int ans4 = 0;
	vector<bool> flag4(n, false);
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			if(red[j].x < blue[i].x && red[j].y < blue[i].y && flag4[j] != true)
			{
				ans4++;
				flag4[j] = true;
				break;
			}
		}
	}
	cout << ((ans1 < ans2 ? ans2 : ans1) < (ans3 < ans4 ? ans4 : ans3) ? (ans3 < ans4 ? ans4 : ans3) : (ans1 < ans2 ? ans2 : ans1))<< endl;
	return 0;
}