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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.io.FileInputStream
import java.util.*


var sc: Scanner = Scanner(System.`in`)
fun main(args: Array<String>) {
    if (args.size > 0) {
        System.setIn(FileInputStream(args[0]))
    }
    sc = Scanner(System.`in`)

    problem()

}

fun problem() {

    val N = sc.nextInt()
    sc.nextLine()
    val reds = mutableListOf<Pair<Int, Int>>()
    for (a in 1..N) {
        val x = sc.nextInt()
        val y = sc.nextInt()
        reds.add(x to y)
    }

    reds.sortBy { it.second }
    reds.sortBy { it.first }
    reds.reverse()
    //println(reds)

    val brues = mutableListOf<Pair<Int, Int>>()
    for (a in 1..N) {
        val x = sc.nextInt()
        val y = sc.nextInt()
        brues.add(x to y)
    }

    brues.sortBy { it.second }
    brues.sortBy { it.first }
    reds.reverse()


    //println(brues)
    println(search(reds, brues, 0))

}


fun search(reds: MutableList<Pair<Int, Int>>, brues: MutableList<Pair<Int, Int>>, count: Int): Int {
    if (reds.size == 0 || brues.size == 0) {
        return count
    }
    var maxcount = count
    for (i in 0..(reds.size - 1)) {
        var red = reds[i]
        for (j in 0..(brues.size - 1)) {

            var brue = brues[j]
            if (red.first < brue.first && red.second < brue.second) {
                reds.removeAt(i)
                brues.removeAt(j)

                val newcount = search(reds, brues, count + 1)
                if (newcount > maxcount) {
                    maxcount = newcount
                }
                reds.add(i, red)
                brues.add(j, brue)
            }
            else {
                break
            }
        }
    }
    return maxcount
}