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 (2.11.7) (judge verdict: Wrong Answer)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val N = in.next().toInt
  val s = new Array[(Int, Int)](N)
  val l = new Array[(Int, Int)](N)

  for(i <- 0 until N) {
    val a, b = in.next().toInt
    s(i) = (a, b)
  }

  for(i <- 0 until N) {
    val c, d = in.next().toInt
    l(i) = (c, d)
  }

  val V = N*2
  val G = Array.fill[List[Int]](V)(Nil) // グラフの隣接リスト表現
  val matching = Array.fill[Int](V)(-1) // マッチングのペア
  val used = new Array[Boolean](V)      // DFSで既に調べたかのフラグ

  // uとvを結ぶ辺をグラフに追加する
  def addEdge(u: Int, v: Int): Unit = {
    G(u) = v :: G(u)
    G(v) = u :: G(v)
  }

  for(a <- 0 until N) {
    val (sx, sy) = s(a)
    for(b <- 0 until N) {
      val (lx, ly) = l(b)
      if (sx < lx && sy < ly)
        addEdge(a, b)
    }
  }

  // 増加パスをDFSで探す
  def dfs(v: Int): Boolean = {
    used(v) = true
    for(i <- 0 until G(v).length) {
      val u = G(v)(i)
      val w = matching(u)
      if(w < 0 || (!used(w) && dfs(w))) {
        matching(v) = u
        matching(u) = v
        return true
      }
    }
    return false
  }

  // 二部グラフの最大マッチングを求める
  def bipartiteMatching(): Int = {
    var res = 0
    for(v <- 0 until V) {
      if(matching(v) < 0) {
        for(i <- 0 until V)
          used(i) = false
        if(dfs(v)) {
          res += 1
        }
      }
    }

    res
  }

  println(bipartiteMatching())
}


class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}