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_imports)]
use std::cmp::{max, min, Ordering};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet, BinaryHeap, VecDeque, BTreeSet, BTreeMap};
#[allow(unused_imports)]
use std::iter::FromIterator;
#[allow(unused_imports)]
use std::io::stdin;

mod util {
    use std::io::stdin;
    use std::str::FromStr;
    use std::fmt::Debug;

    #[allow(dead_code)]
    pub fn line() -> String {
        let mut line: String = String::new();
        stdin().read_line(&mut line).unwrap();
        line.trim().to_string()
    }

    #[allow(dead_code)]
    pub fn gets<T: FromStr>() -> Vec<T>
    where
        <T as FromStr>::Err: Debug,
    {
        let mut line: String = String::new();
        stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|t| t.parse().unwrap())
            .collect()
    }
}

#[allow(unused_macros)]
macro_rules! get {
    ($t:ty) => {
        {
            let mut line: String = String::new();
            stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
                    get!($t)
                   ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
                    get!($($t),*)
                   ).collect::<Vec<_>>()
    };
    ($t:ty ;;) => {
        {
            let mut line: String = String::new();
            stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

fn main() {
    let (n, k) = get!(usize, usize);
    let xy = get!(i64, i64; n);
    let mut x_order = xy.clone();
    let mut y_order = xy.clone();

    x_order.sort_by_key(|xy| xy.0);
    y_order.sort_by_key(|xy| xy.1);

    let mut ans = i64::max_value();

    let mut deq = VecDeque::with_capacity(256);
    for i1 in 0..n {
        for i2 in i1 + 1..n {
            deq.clear();
            let l = x_order[i1].0;
            let r = x_order[i2].0;
            for &(x, y) in &y_order {
                if l <= x && x <= r {
                    deq.push_back(y);

                    while deq.len() > k {
                        deq.pop_front();
                    }

                    if deq.len() == k {
                        ans = min(ans, (r - l) * (deq.back().unwrap() - deq.front().unwrap()));
                    }
                }
            }
        }
    }

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