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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 99999999999999999
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


#define NUM 210

struct LOC{
	int x,y;
};

int V; //頂点数
vector<int> G[NUM]; //グラフの隣接リスト表現
int match[NUM]; //マッチングのペア
bool used[NUM]; //DFSですでに調べたかのフラグ

LOC red[100],blue[100];

//fromとtoを結ぶ辺をグラフに追加する
void add_edge(int from,int to){
	G[from].push_back(to);
	G[to].push_back(from);
}

//増加パスをDFSで探す(node_idのペアを探す)
int dfs(int node_id){
	used[node_id] = true;

	for(int i = 0; i < G[node_id].size(); i++){
		int adj_node_id = G[node_id][i],pair_id = match[adj_node_id];
		if((pair_id < 0)|| //隣接するノードに、まだペアが存在しない
				(used[pair_id] == false && dfs(pair_id) == true)){

			match[node_id] = adj_node_id;
			match[adj_node_id] = node_id;
			return true;
		}
	}
	return false;
}


int bipartie_matching(){
	int ret = 0;
	for(int i = 0; i < V; i++)match[i] = -1; //ペアなしにしておく
	for(int node_id = 0; node_id < V; node_id++){
		if(match[node_id] < 0){ //node_idにペアがいない場合
			for(int i = 0; i < V; i++)used[i] = false;
			if(dfs(node_id)){ //増加パスが存在する場合
				ret++;
			}
		}
	}
	return ret;
}


int main(){

	int N;
	scanf("%d",&N);

	for(int i = 0; i < N; i++){

		scanf("%d %d",&red[i].x,&red[i].y);
	}

	for(int i = 0; i < N; i++){

		scanf("%d %d",&blue[i].x,&blue[i].y);
	}

	for(int i = 0; i < N; i++){
		for(int k = 0; k < N; k++){
			if(red[i].x < blue[k].x && red[i].y < blue[k].y){

				add_edge(i,N+k);
			}
		}
	}

	V = 2*N;

	printf("%d\n",bipartie_matching());

	return 0;
}