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 kotlin.comparisons.compareBy
data class XY(var x: Int, var y: Int) {
    constructor(xy: List<Int>): this(xy[0], xy[1])
}
fun main(args: Array<String>) {
    val N = nextInt()
    val R = Array(N) { XY(listOfInt()) }.sortedWith(compareBy({ -it.y }, { it.x }))
    val B = Array(N) { XY(listOfInt()) }.sortedWith(compareBy({  it.x }, { it.y }))
    var ans = 0
    for (b in 0 until N) {
        var rr = -1
        var maxY = -1
        for (r in 0 until N) {
            if (R[r].x == -1) continue
            if (R[r].x < B[b].x && R[r].y < B[b].y) {
                if (maxY < R[r].y) {
                    rr = r
                    maxY = R[r].y
                }
            }
        }
        if (rr != -1) {
            R[rr].x = -1
            ans += 1
        }
    }
    println(ans)
}

fun next() = readLine()!!
fun nextInt() = next().toInt()
fun listOfString() = next().split(" ")
fun listOfInt(d: Int = 0) = listOfString().map { it.toInt() + d }
fun listOfLong(d: Long = 0) = listOfString().map { it.toLong() + d }