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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;

fn read_line() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s.trim().to_owned()
}

fn read_pair() -> (i64, i64) {
    let s = read_line();
    let mut it = s.split_whitespace().map(|s| s.parse().unwrap());
    let h = it.next().unwrap();
    let w = it.next().unwrap();
    (h, w)
}

fn main() {
    let (n, k) = read_pair();
    let ps: Vec<_> = (0..n).map(|_| read_pair()).collect();

    let mut ans = i64::max_value();

    let (mut xs, mut ys): (Vec<_>, Vec<_>) = ps.iter().cloned().unzip();
    xs.sort();
    ys.sort();

    for ix1 in 0..xs.len() {
        for ix2 in (ix1 + 1)..xs.len() {
            for iy1 in 0..ys.len() {
                for iy2 in (iy1 + 1)..ys.len() {
                    let x1 = xs[ix1];
                    let x2 = xs[ix2];
                    let y1 = ys[iy1];
                    let y2 = ys[iy2];

                    let mut cnt = 0;
                    for &(x, y) in ps.iter() {
                        if x1 <= x && x <= x2 && y1 <= y && y <= y2 {
                            cnt += 1;
                            if cnt >= k {
                                break;
                            }
                        }
                    }

                    if cnt >= k {
                        ans = std::cmp::min(ans, (x2 - x1) * (y2 - y1));
                    }
                }
            }
        }
    }

    println!("{}", ans);
}