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)

//ProblemC

#ifdef _MYPC
#include "mypc.h"
#endif // _MYPC

#include <iostream>
#include <string>
#include <vector>
#include <bitset>
#include <list>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <stack>
#include <algorithm>
#include <numeric>
#include <functional>
#include <memory>
#include <complex>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

//ProblemC
#define MAXN 105

class Solve {
private:
	int n;
	using XY = pair<int,int>;
	int a[MAXN], b[MAXN], c[MAXN], d[MAXN];

	int line[MAXN][MAXN]{};
	int next[MAXN]{}, used[MAXN]{};

	bool found(int x)
	{
		for (int i = 1; i <= n; i++)
		{
			if (line[x][i] && !used[i])
			{
				used[i] = 1;
				if (next[i] == 0 || found(next[i]))
				{
					next[i] = x;
					return 1;
				}
			}
		}
		return 0;
	}

	void solve() {
		cin >> n;
		for (int i = 1; i <= n; i++) cin >> a[i] >> b[i];
		for (int i = 1; i <= n; i++) cin >> c[i] >> d[i];

		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++) {
				if (a[i] < c[j] || b[i] < d[j]) {
					line[i][j] = 1;
				}
			}

		int sum = 0;
		for (int i = 1; i <= n; i++)
		{
			memset(used, 0, sizeof(used));
			if (found(i)) sum++;
		}
		cout << sum << endl;
	}

public:
	Solve() { solve(); }
};

int main(int argc, char* argv[]) {
	unique_ptr<Solve> solve = make_unique<Solve>();
	return 0;
}