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(arg: Array<String>) {
    val (N, K) = listOfInt()
    val XY = Array(N) { listOfLong() }
    var ans = Long.MAX_VALUE
    for (bit in 0L until 1L.shl(N)) {
//        if (java.lang.Long.toBinaryString(bit).count { it == '1'} < K) continue
        var c = 0
        for (n in 0 until N) if (bit.shr(n).and(1) > 0) c += 1
        if (c < K) continue
        var x0 = Long.MAX_VALUE
        var y0 = Long.MAX_VALUE
        var x1 = Long.MIN_VALUE
        var y1 = Long.MIN_VALUE
        for (n in 0 until N) {
            if (bit.shr(n).and(1) > 0) {
                x0 = Math.min(x0, XY[n][0])
                y0 = Math.min(y0, XY[n][1])
                x1 = Math.max(x1, XY[n][0])
                y1 = Math.max(y1, XY[n][1])
            }
        }
        ans = Math.min(ans, (x1 - x0) * (y1 - y0))
    }
    println(ans)
}

fun next() = readLine()!!
fun nextInt(delta: Int = 0) = Integer.parseInt(next()) + delta
fun listOfString() = next().split(" ")
fun listOfInt(d: Int = 0) = listOfString().map { Integer.parseInt(it) + d }
fun listOfLong() = listOfString().map(String::toLong)
fun listOfDouble() = listOfString().map { it.toDouble() }