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>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const double EPS = 1e-10;
const ll INF = 100000000;
const ll MOD = 1000000007;

const int MAX_V = 300;

struct edge{int to, cap, rev; };

vector<edge> G[MAX_V];
int level[MAX_V];
int iter[MAX_V];

void add_edge(int from, int to, int cap){
	G[from].push_back((edge){to, cap, G[to].size()});
	G[to].push_back((edge){from, 0, G[from].size()-1});
}

void bfs(int s){
	memset(level,-1,sizeof(level));
	queue<int> que;
	level[s] = 0;
	que.push(s);
	while(!que.empty()){
		int v = que.front(); que.pop();
		for(int i = 0; i < G[v].size(); i++){
			edge &e = G[v][i];
			if(e.cap > 0 && level[e.to] < 0){
				level[e.to] = level[v]+1;
				que.push(e.to);
			}
		}
	}
}

int dfs(int v, int t, int f){
	if(v == t) return f;
	for(int &i = iter[v]; i < G[v].size(); i++){
		edge &e = G[v][i];
		if(e.cap > 0 && level[v] < level[e.to]){
			int d = dfs(e.to, t, min(f,e.cap));
			if(d > 0){
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}

int max_flow(int s, int t){
	int flow = 0;
	for(;;){
		bfs(s);
		if(level[t] < 0) return flow;
		memset(iter,0, sizeof(iter));
		int f;
		while((f = dfs(s, t, INF)) > 0){
			flow += f;
		}
	}
}

int n;
int a[100], b[100];
int c[100], d[100];

int main() {
    cin >> n;
    rep(i,n) cin >> a[i] >> b[i];
    rep(i,n) cin >> c[i] >> d[i];
    rep(i,n) add_edge(2*n, i, 1);
    rep(i,n) add_edge(n+i, 2*n+1, 1);
    rep(i,n) {
        rep(j,n) {
            if (a[i] < c[j] && b[i] < d[j]) {
                add_edge(i, j+n, 1);
            }
        }
    }
    cout << max_flow(2*n, 2*n+1) << endl;
}