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: Wrong Answer)

import java.util.Collections;
import java.util.Scanner;
import java.util.ArrayList;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();

    int[][] red = new int[n][2];
    int[][] blue = new int[n][2];

    for (int i = 0; i < n; ++i) {
      red[i][0] = sc.nextInt();
      red[i][1] = sc.nextInt();
    }

    for (int i = 0; i < n; ++i) {
      blue[i][0] = sc.nextInt();
      blue[i][1] = sc.nextInt();
    }

    ArrayList<int[]> redL = new ArrayList<>();
    Collections.addAll(redL, mergeSort(red));

    ArrayList<int[]> blueL = new ArrayList<>();
    Collections.addAll(blueL, mergeSort(blue));
    int res = 0;

    for (int j = redL.size() - 1; j >= 0; --j) {
      int[] rPoint = redL.get(j);
      for (int i = blueL.size() - 1; i >= 0; --i) {
        int[] bPoint = blueL.get(i);
        if (rPoint[0] < bPoint[0] && rPoint[1] < bPoint[1]) {
          ++res;
          blueL.remove(i);
          break;
        }
      }
    }


    System.out.println(res);
  }

  public static int[][] mergeSort(int[][] a) {
    if (a.length <= 1)
      return a;

    int mid = a.length / 2;
    int[][] left = new int[mid][2];
    int[][] right = new int[a.length - mid][2];

    for (int i = 0; i < mid; ++i)
      left[i] = a[i];

    for (int i = mid; i < a.length; ++i)
      right[i - mid] = a[i];

    return merge(mergeSort(left), mergeSort(right));
  }

  public static int[][] merge (int[][] l, int[][] r) {
    int[][] merged = new int[l.length + r.length][2];

    int i = 0, j = 0, k = 0;
    while (i < l.length && j < r.length) {
      if (l[i][0] < r[j][0] || (l[i][0] == r[j][0] && l[i][1] <= r[i][1]))
        merged[k++] = l[i++];
      else
        merged[k++] = r[j++];
    }

    while (i < l.length)
      merged[k++] = l[i++];
    while (j < r.length)
      merged[k++] = r[j++];

    return merged;
  }
}