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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
class bipartite_matching {
public:
    int n;
    std::vector<std::vector<int>> g;
    std::vector<int> match;

    bipartite_matching(int n_) : n(n_), g(n_), match(n_), used(n_) {}

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

    int maximum_matching(void) {
        int res = 0;
        fill(begin(match), end(match), -1);
        for (int v = 0; v < n; ++v) {
            if (match[v] == -1) {
                fill(begin(used), end(used), false);
                if (dfs(v)) res++;
            }
        }
        return res;
    }

private:
    std::vector<int> used;
    bool dfs(int v) {
        used[v] = true;
        for (int u : g[v]) {
            int w = match[u];
            if (w == -1 || (!used[w] && dfs(w))) {
                match[v] = u;
                match[u] = v;
                return true;
            }
        }
        return false;
    }
};
bool judge(P r, P b) {
    return ((r.first < b.first) && (r.second < b.second));
}
int main() {
    int n; cin >> n;
    bipartite_matching bm(2*n);
    vector<P> red(n);
    rep(i, n) {
        cin >> red[i].first >> red[i].second;
    }
    vector<P> blu(n);
    rep(i, n) {
        cin >> blu[i].first >> blu[i].second;
    }
    rep(i, n) {
        rep(j, n) {
            if(judge(red[i], blu[j])) {
                bm.add_edge(i, n+j);
            }
        }
    }
    cout << bm.maximum_matching() << endl;
}