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>
using namespace std;

#define rep(i, n) for (int (i) = 0 ; (i) < (int)(n) ; ++(i))
#define repn(i, m, n) for (int (i) = (m) ; (i) < (int)(n) ; ++(i))
#define rep_rev(i, n) for (int (i) = (int)(n) - 1 ; (i) >= 0 ; --(i))
#define repn_rev(i, m, n) for (int (i) = (int)(n) - 1 ; (i) >= (m) ; --(i))

#define INF 2e9
#define ll long long
#define MOD 1e9+7
#define MAX_V 20005

/* 辺を表す構造体 */
struct edge {int to, cap ,rev; };

/* グラフの隣接リスト表現 */
vector<edge> G[MAX_V];
/* DFSですべに調べたかのグラフ */
bool used[MAX_V];

/* fromからtoへ向かう容量capの辺をグラフに追加する */
/* 相互参照できるようにしている */
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});    
}

/* 増加パスをdfsで探す */
int dfs(int v, int t, int f){
    if(v==t) return f;
    used[v] = true;
    for(int i=0; i<G[v].size(); i++){
        edge &e = G[v][i];
        if(!used[e.to] && e.cap>0){
            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(;;){
        memset(used, 0, sizeof(used));
        int f = dfs(s, t, INF);
        if(f==0) return flow;
        flow += f;
    }
}

int main() {
    int N;
    cin >> N;
    vector<int> a(N), b(N), c(N), d(N);
    rep(i, N) cin >> a[i] >> b[i];
    rep(i, N) cin >> c[i] >> d[i];

    /* 0: 始点, 1~N: 赤, N+1~2N: 青, 2N+1: 終点 */
    rep(i, N) {
        add_edge(0, i+1, 1);
        add_edge(i+N+1, 2*N+1, 1);
    }

    rep(i, N) rep(j, N) if(a[i] < c[j] && b[i] < d[j]) add_edge(i+1, j+N+1, 1);
    
    cout << max_flow(0, 2*N+1) << endl;

    return 0;
}