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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Scanner;
import java.util.Comparator;
import java.util.Collections;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Kenji
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ABC091C solver = new ABC091C();
        solver.solve(1, in, out);
        out.close();
    }

    static class ABC091C {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int N = in.nextInt();
            int[] redX = new int[N];
            int[] redY = new int[N];
            int[] blueX = new int[N];
            int[] blueY = new int[N];
            boolean[] used = new boolean[N];
            int ans = 0;
            List<Integer> list = new ArrayList<>();
            for (int i = 0; i < N; i++) {
                redX[i] = in.nextInt();
                redY[i] = in.nextInt();
                used[i] = false;
            }
            for (int i = 0; i < N; i++) {
                blueX[i] = in.nextInt();
                blueY[i] = in.nextInt();
                list.add(i);
            }

            // sort
            Collections.sort(list, Comparator.comparingInt(o -> blueX[o]));

            for (Integer i : list) {
                List<Integer> candidate = new ArrayList<>();
                for (int j = 0; j < N; j++) {
                    if (redX[j] < blueX[i]
                            && redY[j] < blueY[i]
                            && !used[j]) {
                        candidate.add(j);
                    }
                }

                double maxY = -1;
                int index = -1;
                for (Integer j : candidate) {
                    double d = redY[j];
                    if (maxY < d) {
                        maxY = d;
                        index = j;
                    }
                }
                if (index != -1) {
                    ans++;
                    used[index] = true;
                }
            }
            out.println(ans);
        }

    }
}