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: Accepted)

case class pos(val x: Int, val y: Int)

object Main extends App {
	val n = scala.io.StdIn.readInt
	var red = new Array[pos](n)
	var blue = new Array[pos](n)

	for(i<-0 until n){
		val temp = scala.io.StdIn.readLine.split(" ").map(_.toInt)
		red(i) = pos(temp(0),temp(1))
	}

	for(i<-0 until n){
		val temp = scala.io.StdIn.readLine.split(" ").map(_.toInt)
		blue(i) = pos(temp(0),temp(1))
	}

	red = red.sortBy(_.x)
	blue = blue.sortBy(_.x)

	var ans = 0
	for(x <- blue){
		var idx = -1
		var max = -1

		for(i<-0 until n){
			if(red(i).x != -1){
				if(red(i).x <= x.x){
					if(red(i).y > max && red(i).y < x.y){
					idx = i
					max = red(i).y
				}
				}
			}
		}

		if(idx != -1){
			red(idx) = pos(-1,-1)
			ans += 1
		}
	}

	println(ans)

}