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)

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <fstream>
#include <sstream>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <queue>
#include <bitset>     //UWAGA - w czasie kompilacji musi byc znany rozmiar wektora - nie mozna go zmienic
#include <cassert>
#include <iomanip>        //do setprecision
#include <ctime>
#include <complex>
using namespace std;

#define FOR(i,b,e) for(int i=(b);i<(e);++i)
#define FORQ(i,b,e) for(int i=(b);i<=(e);++i)
#define FORD(i,b,e) for(int i=(b)-1;i>=(e);--i)
#define REP(x, n) for(int x = 0; x < (n); ++x)

#define ST first
#define ND second
#define PB push_back
#define MP make_pair
#define LL long long
#define ULL unsigned LL
#define LD long double

const double pi = 3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
const int mod=1000000007;
typedef pair<int,int> pii;
typedef pair<int,pair<int,int>> piii;

int V;//頂点数
const int SIZE_OF_G=400;
vector<int> G[SIZE_OF_G];
int match[SIZE_OF_G];
bool used[SIZE_OF_G];

void init(int v){
  V=v;
  for(int i=0;i<V;i++)G[i].clear();
}

void add_edge(int u,int v){
  G[u].push_back(v);
  G[v].push_back(u);
}

bool dfs(int v){
  used[v]=true;
  for(int i=0; i<G[v].size();i++){
    int u=G[v][i],w=match[u];
    if(w<0||(!used[w]&&dfs(w))){
      match[v]=u;
      match[u]=v;
      return true;
    }
  }
  return false;
}

int bipartite_match(){
  int res=0;
  memset(match,-1,sizeof(match));
  for(int v=0; v<V;v++){
    if(match[v]<0){
      memset(used,0,sizeof(used));
      if(dfs(v))res++;
    }
  }
  return res;
}

int main(){
	int n;
	pii x[211]={},y[211]={};//(0:無,1:赤、２：青),idx
	int a[111]={},b[111]={},c[111]={},d[111]={};
	cin>>n;
	V=n;
	int ans=0;
	FOR(i,0,n){
		cin>>a[i]>>b[i];
		x[a[i]]=make_pair(1,i);
		y[b[i]]=make_pair(1,i);
	}
	FOR(i,0,n){
		cin>>c[i]>>d[i];
		x[c[i]]=make_pair(2,i);;
		y[d[i]]=make_pair(2,i);;
	}
	FOR(i,0,n){
		FOR(j,0,n){
			if(a[i]<c[i]&&b[i]<d[i])add_edge(i,j+n);
		}
	}

	cout<<bipartite_match()<<endl;
    return 0;
}