Problem:
Score : 400 points

Problem Statement

We have N points in a two-dimensional plane.

The coordinates of the i-th point (1 \leq i \leq N) are (x_i,y_i).

Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior.

Here, points on the sides of the rectangle are considered to be in the interior.

Find the minimum possible area of such a rectangle.

Constraints

2 \leq K \leq N \leq 50

-10^9 \leq x_i,y_i \leq 10^9 (1 \leq i \leq N)

x_i≠x_j (1 \leq i<j \leq N)

y_i≠y_j (1 \leq i<j \leq N)

All input values are integers. (Added at 21:50 JST)

Input

Input is given from Standard Input in the following format:

N K
x_1 y_1
:
x_{N} y_{N}

Output

Print the minimum possible area of a rectangle that satisfies the condition.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val (n, k) = readLine()!!.split(' ').map { it.toInt() }
    val xs = IntArray(n)
    val ys = IntArray(n)
    for (i in 0 until n) {
        val (x, y) = readLine()!!.split(' ').map { it.toInt() }
        xs[i] = x
        ys[i] = y
    }
    xs.sort()
    ys.sort()

    var ans = 1L * (xs.last() - xs.first()) * (ys.last() - ys.first())

    for (xi in 0 until n) {
        for (xj in xi + 1 until n) {
            for (yi in 0 until n) {
                for (yj in yi + 1 until n) {
                    val xl = xs[xi]
                    val xr = xs[xj]
                    val yb = ys[yi]
                    val yt = ys[yj]

                    val cnt = (0 until n).count { xs[it] in xl..xr && ys[it] in yb..yt }
                    if (cnt >= k) {
                        ans = Math.min(1L * (xr - xl) * (yt - yb), ans)
                    }
                }
            }
        }
    }

    println(ans)
}