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)

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

#define MX 205
pair<int, int> a[MX], b[MX];
vector<int> g[MX];
int lnk[MX];
bool vis[MX];

bool dfs(int u) {
	for (int v : g[u]) {
		if (vis[v]) continue;
		vis[v] = true;
		if (lnk[v] == -1 || dfs(lnk[v])) {
			lnk[v] = u;
			return 1;
		}
	}
	return 0;
}

int main() {
	int n; scanf("%d", &n);
	for(int i = 0; i < n; i++) scanf("%d%d", &a[i].first, &a[i].second);
	for(int i = 0; i < n; i++) scanf("%d%d", &b[i].first, &b[i].second);
	for(int i = 0; i < n; i++)
		for(int j = 0; j < n; j++)
			if(a[i].first < b[j].first && a[i].second < b[j].second) {
				g[i].push_back(j + n);
				g[j + n].push_back(i);
			}
	for(int i = n; i < n + n; i++) lnk[i] = -1;
	int res = 0;
	for(int i = 0; i < n; i++) {
		for(int j = n; j < n + n; j++) vis[j] = 0;
		res += dfs(i);
	}
	printf("%d\n", res);
	return 0;
}