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)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
    ($iter:expr, mut $var:ident $($r:tt)*) => {
        let mut $var = $iter.next().unwrap().parse::<usize>().unwrap();
        input_inner!{$iter $($r)*}
    };
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
    ($iter:expr, $var:ident $($r:tt)*) => {
        let $var = $iter.next().unwrap().parse::<usize>().unwrap();
        input_inner!{$iter $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().unwrap()
    };
}
#[allow(unused_macros)]
macro_rules! outbuf {
    ($var:ident) => {
        use std::io::{stdout, BufWriter, Write};
        let stdout = stdout();
        let mut $var = BufWriter::new(stdout.lock());
    };
}
use std::cmp::min;
fn main() {
    input! {n, k, xy: [(isize, isize); n]};
    // outbuf! {out};
    let mut ans = std::isize::MAX;
    let (mut x, mut y): (Vec<_>, Vec<_>) = xy.iter().cloned().unzip();
    let ok = |x1: isize, x2: isize, y1: isize, y2: isize| {
        let mut cnt = 0;
        for &(x, y) in &xy {
            if x1 <= x && x <= x2 && y1 <= y && y <= y2 {
                cnt += 1;
            }
        }
        cnt
    };
    x.sort();
    y.sort();
    for xi in 0..(n-1) {
        for xj in xi..n {
            let mut yj = 0;
            for yi in 0..n {
                while yj < n && ok(x[xi], x[xj], y[yi], y[yj]) < k {
                    yj += 1;
                }
                if yj == n {
                    yj -= 1;
                }
                if ok(x[xi], x[xj], y[yi], y[yj]) >= k {
                    ans = min(ans, (x[xi]-x[xj]).abs()*(y[yi]-y[yj]).abs());
                }
                if yi == yj {
                    yj += 1;
                }
            }
        }
    }
    println!("{}", ans);
}