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 OCaml (4.02.3) (judge verdict: Accepted)

Scanf.scanf "%d %d" (fun n k ->
    let xy = Array.init n (fun _ -> Scanf.scanf " %d %d" (fun x y -> x, y)) in
    let x, y = Array.fold_left (fun (xl, yl) (x, y) -> x :: xl, y :: yl) ([], []) xy in
    let x = Array.of_list x in
    let y = Array.of_list y in
    Array.sort compare x;
    Array.sort compare y;
    let rec loop1 i1 acc =
        let rec loop2 i2 x1 acc =
            let rec loop3 i3 x2 acc =
                let rec loop4 i4 y1 acc =
                    let count y2 =
                        Array.fold_left (fun acc (x, y) -> if x >= x1 && x <= x2 && y >= y1 && y <= y2 then acc + 1 else acc) 0 xy
                    in
                    if i4 >= n then acc else
                        let kk = count y.(i4) in
                        let acc = if kk >= k then min acc ((x2 - x1) * (y.(i4) - y1)) else acc  in
                        loop4 (i4 + 1) y1 acc
                in
                if i3 >= n - 1 then acc else loop3 (i3 + 1) x2 (loop4 (i3 + 1) y.(i3) acc)
            in
            if i2 >= n then acc else loop2 (i2 + 1) x1 (loop3 0 x.(i2) acc)
        in
        if i1 >= n - 1 then acc else loop1 (i1 + 1) (loop2 (i1 + 1) x.(i1) acc)
    in
    loop1 0 max_int |> Printf.printf "%d\n"
)