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: Runtime Error)

object Main extends App{
  val s=new java.util.Scanner(System.in)
  val n=s.nextInt()
  def f=Iterator.continually{(s.nextInt,s.nextInt)}
  val (xs,ys)=f.take(n*2).toVector.splitAt(n)
  val l=Array.ofDim[(Int,(Int, Int),(Int,Int))](n,n)
  for(i<-0 until n;j<-0 until n){
    val (a,b) = xs(i)
    val (c,d) = ys(j)
    if(a<c&&b<d)l(i)(j)=(c-a+d-b,(a,b),(c,d))
  }
  val p=Array.ofDim[Int](100,100)
  l.toSeq.flatten.filter(_!=null).sortBy(_._1).foreach{case (_,(a,b),(c,d))=>
      if(p(a)(b)==0&&p(c)(d)==0) {
        p(a)(b)=1
        p(c)(d)=1
      }
  }
  println(p.map(_.sum).sum/2)
}