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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <random>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
using namespace std;

const int INF = 1 << 30;

struct Edge {
    int to; int cap; int rev;
    Edge(int t, int c, int r) : to(t), cap(c), rev(r) {}
};

struct Flow {
    vector <vector<Edge>> G;
    vector<bool> used;
    Flow(int n) : G(n), used(n, false) {}
    void add_edge(int from, int to, int cap) {
        G[from].push_back(Edge(to, cap, G[to].size()));
        G[to].push_back(Edge(from, 0, G[from].size() - 1));
    }
    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;
        while (1) {
            fill(used.begin(), used.end(), false);
            int f = dfs(s, t, INF);
            if (f == 0) return flow;
            flow += f;
        }
    }
};

int main() {
    int n;
    cin >> n;

    Flow fl(n*2+2);
    int a[n], b[n], c[n], d[n];
    for (int i = 0; i < n; i++) {
        cin >> a[i] >> b[i];
        fl.add_edge(n*2, i, 1);
    }
    for (int i = 0; i < n; i++) {
        cin >> c[i] >> d[i];
        fl.add_edge(n+i, n*2+1, 1);
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (a[i] < c[j] && b[i] < d[j]) {
                fl.add_edge(i, n+j, 1);
            }
        }
    }

    cout << fl.max_flow(n*2, n*2+1) << "\n";
    return 0;
}