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::ops::{Index, IndexMut};
// use std::cmp::{Ordering, min, max};
// use std::collections::{BinaryHeap, BTreeMap};
// use std::collections::btree_map::Entry::{Occupied, Vacant};
// use std::clone::Clone;

fn getline() -> String{
    let mut res = String::new();
    std::io::stdin().read_line(&mut res).ok();
    res
}

macro_rules! readl {
    ($t: ty) => {
        {
            let s = getline();
            s.trim().parse::<$t>().unwrap()
        }
    };
    ($( $t: ty),+ ) => {
        {
            let s = getline();
            let mut iter = s.trim().split(' ');
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*) 
        }
    };
}

macro_rules! readlvec {
    ($t: ty) => {
        {
            let s = getline();
            let iter = s.trim().split(' ');
            iter.map(|x| x.parse().unwrap()).collect::<Vec<$t>>()
        }
    }
}

macro_rules! mvec {
    ($v: expr, $s: expr) => {
        vec![$v; $s]
    };
    ($v: expr, $s: expr, $($t: expr),*) => {
        vec![mvec!($v, $($t),*); $s]
    };
}

macro_rules! debug {
    ($x: expr) => {
        println!("{}: {:?}", stringify!($x), $x)
    }
}

fn printiter<'a, T>(v: &'a T)
where
    &'a T: std::iter::IntoIterator, 
    <&'a T as std::iter::IntoIterator>::Item: std::fmt::Display {
    for (i,e) in v.into_iter().enumerate() {
        if i != 0 {
            print!(" ");
        }
        print!("{}", e);
    }
    println!("");
}

fn main() {
    let (n, k) = readl!(usize, usize);
    let mut x = Vec::with_capacity(n);
    let mut y = Vec::with_capacity(n);
    for _ in 0..n {
        let (a, b) = readl!(i64, i64);
        x.push(a);
        y.push(b);
    }
    let x = x;
    let y = y;
    
    let mut ans = std::i64::MAX;
    for lx in &x {
        for rx in &x {
            if rx <= lx {
                continue;
            }
            for ly in &y {
                for ry in &y {
                    if ry <= ly {
                        continue;
                    }
                    let mut cnt = 0;
                    for i in 0..n {
                        if *lx <= x[i] && x[i] <= *rx
                        && *ly <= y[i] && y[i] <= *ry {
                            cnt += 1;
                        }
                    }
                    if cnt >= k {
                        ans = std::cmp::min((rx-lx)*(ry-ly), ans);
                    }
                }
            }
        }
    }
    println!("{}", ans);
}