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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
 
template <typename Weight, int MAX_N, Weight INF> class MaxFlow{
private:
	struct E{
		int to, rev;
		Weight lim;
		E(int x, Weight y, int z){
			to = x;
			lim = y;
			rev = z;
		}
	};
 
	vector<E> edge[MAX_N];
	int lev[MAX_N];
	int iter[MAX_N];
 
 
	Weight dfs(int s, int g, Weight f){
		if(s == g)return f;
		for(int &i = iter[s];i < edge[s].size();i++){
			E &e = edge[s][i];
			if(lev[e.to] <= lev[s] || e.lim <= 0)continue;
			Weight tmp = dfs(e.to, g, min(f, e.lim));
			if(tmp<1e-9)continue;
			e.lim -= tmp;
			edge[e.to][e.rev].lim += tmp;
			return tmp;
		}
		return 0;
	}
 
	void bfs(int x){
		queue<int> q;
		q.push(x);
		int p = 0;
		while(!q.empty()){
			for(int i = q.size();i--;){
				int tmp = q.front();q.pop();
				if(lev[tmp] != MAX_N)continue;
				lev[tmp] = p;
				for(int j = 0;j < edge[tmp].size();j++){
					if(edge[tmp][j].lim > 0 && lev[edge[tmp][j].to] == MAX_N)q.push(edge[tmp][j].to);
				}
			}
			p++;
		}
	}
 
public:
	void make_edge(int a, int b, Weight l){
		edge[a].push_back(E(b, l, edge[b].size()));
		edge[b].push_back(E(a, 0, edge[a].size() - 1));
	}
 
	Weight max_flow(int s, int g){
		Weight res = 0;
		bool fin = false;
		while(!fin){
			fill(lev, lev + MAX_N, MAX_N);
			fill(iter, iter + MAX_N, 0);
			bfs(s);
			fin = true;
			while(true){
				Weight p = dfs(s, g, INF);
				if(p < 1e-9)break;
				fin = false;
				res += p;
			}
		}	
		return res;
	}
};

int n;
int a[108000], b[108800], c[10800], d[10880];

int main(){

	MaxFlow<int, 216, 108000> mf;	
	cin >> n;
	for(int i = 0;i < n;i++)cin >> a[i] >> b[i];
	for(int i = 0;i < n;i++)cin >> c[i] >> d[i];
	int start = 213, goal = 214;
	
	for(int i = 0;i < n;i++){
		mf.make_edge(start, i, 1);
		mf.make_edge(i + 100, goal, 1);
		for(int j = 0;j < n;j++){
			if(a[i] < c[j] && b[i] < d[j]){
				mf.make_edge(i, j + 100, 1);
			}
		}
	}
	cout << mf.max_flow(start, goal) << endl;
	return 0;
}