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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) REP(i, 0, n)
#define rrep(i, n) for (int i = (int)(n-1); i >= 0; i--)
#define sz(x) int(x.size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define INF 2e9
#define LINF 1e18
#define mod 1000000007
template<class T> inline bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T &a, const T &b) { if (a > b) { a = b; return 1; } return 0; }
typedef long long ll;
typedef pair<int, int> P;
int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};


int main() {
    int n; cin >> n;
    vector<P> r(n), b(n);
    int x, y;
    rep(i,n) {
        cin >> x >> y;
        r[i].first = x;
        r[i].second = y;
    }
    rep(i,n) {
        cin >> x >> y;
        b[i].first = x;
        b[i].second = y;
    }

    sort(all(r));
    sort(all(b));

    int ans = 0;

    rep(i,n) { //blueの点のループ
        int k = 0;
        rep(j,n) { //red
            if (r[j].first >= b[i].first) break;
            k++;
        }
        int y_mx = -1; int bangou;
        bool exist = false;
        rep(j,k) { //y座標maxをとる
            if (r[j].second < b[i].second) {
                exist = true;
                if (y_mx < r[j].second) {
                    y_mx = r[j].second;
                    bangou = j;
                }
            }
        }
        if (exist) {
            ans++;
            r[bangou].first = 1000; //採用した点の実質削除
            sort(all(r));
        }
    }

    cout << ans << endl;
}