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)

// g++ macro.cpp -std=c++14
#include <bits/stdc++.h>
typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;
using namespace std;

#define dump(x)  cout << #x << " = " << (x) << endl;
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl
#define possible(n) cout << ((n) ? "possible" : "impossible"  ) << endl

#define SANKOU(n,a,b) cout << ((n) ? (#a) : (#b) ) << endl

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)

#define FOREACH(x,a) for(auto& (x) : (a) )

#define WFA(d,v) REP(k,v)REP(i,v)REP(j,v)d[i][j]=min(d[i][j],d[i][k]+d[k][j])

#define SCOUT(x) cout<<(x)<<" "
#define ENDL cout<<endl

#define VECCIN(x) for(auto&youso_: (x) )cin>>youso_
#define VECIN2(x,y) REP(i,x.size())cin>>x[i]>>y[i]
#define VECCOUT(x) if(1){for(auto tt=x.begin();tt!=x.end();tt++){if(tt!=x.begin())cout<<" ";cout<<(*tt);}cout<<endl;}

#define ALL(obj) (obj).begin(),(obj).end()

#define EXIST(n,x) (find(ALL(n),x)!=n.end())
#define UNIQUE(obj) sort(ALL( obj )); obj.erase(unique(ALL(obj)),obj.end())
#define EN(x) if(1){cout<<#x<<endl;return 0;}
#define COUT(x) cout<<(x)<<endl
void CINT(){}
template <class Head,class... Tail>
void CINT(Head&& head,Tail&&... tail){
    cin>>head;
    CINT(move(tail)...);
}
#define CIN(...) int __VA_ARGS__;CINT(__VA_ARGS__)
#define LCIN(...) ll __VA_ARGS__;CINT(__VA_ARGS__)
#define SCIN(...) string __VA_ARGS__;CINT(__VA_ARGS__)

template <class T = ll>
T IN(){T x;cin>>x;return (x);}

template <class Head>
void VT(Head head){}
template <class Head,class Seco,class... Tail>
void VT(Head&& head,Seco&& seco,Tail&&... tail){
    seco.resize(head);
    VT(head,move(tail)...);
}
void VT2(){}
template <class Head,class... Tail>
void VT2(Head&& head,Tail&&... tail){
  VECCIN(head);
  VT2(move(tail)...);
}

template <class Head>
void VT3(Head&& head){}
template <class Head,class Seco,class... Tail>
void VT3(Head&& head,Seco&& seco,Tail&&... tail){
  seco[head]=IN();
  VT3(head,move(tail)...);
}

#define VC1(n,...) V __VA_ARGS__;VT(n,__VA_ARGS__);VT2(__VA_ARGS__); //aaabbbccc
#define VC2(n,...) V __VA_ARGS__;VT(n,__VA_ARGS__);REP(i,n)VT3(i,__VA_ARGS__); //abcabcabc

// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision; // cpp_int

#define P pair<int,int>
#define V vector<int>
#define M map<int,int>
#define S set<int>

#define pb(a) push_back(a)
#define mp make_pair

class FordFulkerson{//最大流(最小カット)を求める 辺の形式に注意
  public:
  FordFulkerson(vector<M> ed,int n);
  int Dfs(int v,int t,int f);
  ll Solve(int s,int t);
  int pnum;
  vector<M> edges;
  V flag;
};

FordFulkerson::FordFulkerson(vector<M> ed,int n){
  pnum=n;
  edges=ed;
}

int FordFulkerson::Dfs(int s,int t,int f){
  if(s==t)return f;
  flag[s]=true;
  FOREACH(e,edges[s]){
    if(!flag[e.first] && e.second>0){
      int d = Dfs(e.first,t,min(f,e.second));
      if(d>0){
        e.second-=d;
        edges[e.first][s]+=d;
        return d;
      }
    }
  }
  return 0;
}

ll FordFulkerson::Solve(int s,int t){
  ll ret=0;
  while(1){
    flag=V(pnum,0);
    int f=Dfs(s,t,INF);
    if(!f)return ret;
    ret+=f;
  }
}


int main(){
  CIN(n);VC2(n,a,b);VC2(n,c,d);
  sort(ALL(a));
  sort(ALL(b));
  sort(ALL(c));
  sort(ALL(d));
  vector<M> ed(2*n+3);//0-1
  V taiou(n);
  V taiou2(n);
  REP(i,n)taiou[i] = i+2;
  REP(i,n)taiou2[i] = i+2+n;
  REP(i,n){
    ed[0][taiou[i]]=1;
    ed[taiou2[i]][1]=1;
  }
  REP(i,n)
    REP(j,n)
      if(a[i]<c[i] && b[i] < d[i]){
        ed[taiou[i]][taiou2[j]]=1;
        ed[taiou2[j]][taiou[i]]=1;
      }
  FordFulkerson calc(ed,2*n+3);
  int ans = calc.Solve(0,1);
  COUT(ans);


  return 0;
}