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: Time Limit Exceeded)

fun main(args: Array<String>) {
    val (n, k) = readLine()!!.split(' ').map(String::toInt)
    val xyList = mutableListOf<Pair<Long, Long>>()
    (0 until n).forEach {
        val (x, y) = readLine()!!.split(' ').map(String::toLong)
        xyList.add(Pair(x, y))
    }
    var ans = Long.MAX_VALUE / 2
    for (i1 in 0 until n) {
        val p1 = xyList[i1]
        for (i2 in 0 until n) {
            val p2 = xyList[i2]
            for (i3 in 0 until n) {
                val p3 = xyList[i3]
                for (i4 in 0 until n) {
                    val p4 = xyList[i4]
                    val minX = Math.min(p1.first, Math.min(p2.first, Math.min(p3.first, p4.first)))
                    val minY = Math.min(p1.second, Math.min(p2.second, Math.min(p3.second, p4.second)))
                    val maxX = Math.max(p1.first, Math.max(p2.first, Math.max(p3.first, p4.first)))
                    val maxY = Math.max(p1.second, Math.max(p2.second, Math.max(p3.second, p4.second)))
                    var countInnerP = 0
                    for (j in 0 until n) {
                        val (x, y) = xyList[j]
                        if (x in minX..maxX && y in minY..maxY) countInnerP++
                    }
                    if (countInnerP >= k) ans = Math.min(ans, (maxX - minX) * (maxY - minY))
                }
            }
        }
    }
    println(ans)
}