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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define rep1(i,n) for(int i=1;i<=(int)n;i++)
#define sp(n) cout << fixed << setprecision(n)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
typedef long long ll;
using namespace std;
const int INF = 1e9;
const int MAX_V= 1e6;
//辺を表す構造体{行先、容量、逆辺}
struct edge{int to, cap, rev; };
vector<vector<edge>> G(MAX_V);//グラフの隣接リスト表現
vector<int> level(MAX_V);//sからの距離
vector<int> iter(MAX_V);//どこまで調べ終わったか

//fromからtoへ向かう容量capの辺をグラフに追加する
void add_edge(int from, int to, int cap){
    G[from].push_back((edge){to,cap,(int)G[to].size()});
    G[to].push_back((edge){from,0,(int)G[from].size()-1});
}

//sからの最短距離をBFSで計算する
void bfs(int s){
    rep(i,MAX_V)level[i]=-1;
    queue<int> que;
    level[s]=0;
    que.push(s);
    while(!que.empty()){
        int v=que.front();que.pop();
        rep(i,G[v].size()){
            edge& e=G[v][i];
            if(e.cap>0&&level[e.to]<0){
                level[e.to]=level[v]+1;
                que.push(e.to);
            }
        }
    }
}

//追加パスをdfsであらわす
int dfs(int v,int t,int f){
    if(v==f)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;
}

//sからtへの最大流を求める
int max_flow(int s,int t){
    int flow=0;
    for(;;){
        bfs(s);
        if(level[t]<0)return flow;
        rep(i,MAX_V)iter[i]=0;
        int f;
        while((f=dfs(s,t,INF))>0){
            flow+=f;
        }
    }
}

int main(void){
    int n;cin>>n;
    vector<pair<int,int>> ps(n),pt(n);
    rep(i,n) cin>>ps[i].first>>ps[i].second;
    rep(i,n) cin>>pt[i].first>>pt[i].second;
    int s=2*n,t=s+1;
    rep(i,n)add_edge(s,i,1);
    rep(i,n)add_edge(n+i,t,1);
    rep(i,n)rep(j,n){
        if(ps[i].first<pt[j].first&&ps[i].second>pt[j].second) add_edge(i,n+j,1);
    }
    //return 0;
    int res=max_flow(s,t);
    cout<<res<<endl;
}