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: Accepted)

import java.util.*
import kotlin.*


fun main(args: Array<String>){
    var line: String = readLine() as String
    val n = line.toInt()
    val red =ArrayList<IntArray>()
    for(i in 0 until n){
        val a=readLine()!!.split(" ")
        red.add(intArrayOf(a[0].toInt(),a[1].toInt()))
    }

    val blue = PriorityQueue<IntArray>(n,Comparator<IntArray>{ o1:IntArray, o2:IntArray -> o1[0]-o2[0] })
    for(i in 0 until n){
        val a=readLine()!!.split(" ")
        blue.add(intArrayOf(a[0].toInt(),a[1].toInt()))
    }

    red.sortBy { it:IntArray->it[1] }//y座標でソート
    var count=0
    while (true){
        val next = blue.poll() ?: break
        var index=red.size-1
        while(index>=0){
            if(red[index][0]<next[0]&&red[index][1]<next[1]){
                count+=1
                red.removeAt(index)
                break
            }
            index -= 1
        }
        if(red.size==0){
            break
        }
    }
    print(count)
}