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.util.Comparator

private fun readStrings() = readLine()!!.split(" ")
private fun readInts() = readStrings().map { it.toInt() }

val pairComparator = Comparator { c1: Pair<Int, Int>, c2: Pair<Int, Int> ->
    val f =  c1.first.compareTo(c2.first)
    if (f == 0) c1.second.compareTo(c2.second) else f
}

fun main(args: Array<String>) {
    val N = readInts()[0]
    val reds = (1..N).map {
        val (a, b) = readInts()
        a to b
    }.sortedWith(pairComparator)
    val blues = (1..N).map {
        val (a, b) = readInts()
        a to b
    }.sortedWith(pairComparator)

    var counts = 0
    var lastBlueIndex = -1
    reds.forEach { red ->
        while (true) {
            lastBlueIndex ++
            if (lastBlueIndex >= N) return@forEach

            if (pairComparator.compare(red, blues[lastBlueIndex]) < 0) {
                counts++
                break
            }
        }
    }

    println(counts)
}