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 Scala (2.13.1) (judge verdict: Accepted)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n = sc.nextInt()
  val reds, blues = List.fill(n)((sc.nextInt(), sc.nextInt()))
  val e = for {
    ((a, b), i) <- reds.zipWithIndex
    ((c, d), j) <- blues.zipWithIndex
    if a < c && b < d
  } yield (i, j)

  println(maximumMatching(n, n, e))

  def maximumMatching(n: Int, m: Int, e: Seq[(Int, Int)]): Int = {
    val adj = Array.fill(n)(List[Int]())
    for ((u, v) <- e) {
      adj(u) ::= v
    }

    val pair = Array.fill(m)(-1)

    def find(u: Int): Boolean = {
      val visited = Array.fill(m)(false)

      def dfs(u: Int): Boolean = {
        adj(u).find { v =>
          if (!visited(v)) {
            visited(v) = true
            pair(v) == -1 || dfs(pair(v))
          } else {
            false
          }
        } match {
          case Some(v) =>
            pair(v) = u
            true
          case None => false
        }
      }

      dfs(u)
    }

    (0 until n).count(find)
  }
}