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::vec::Vec;
use std::cmp::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
  let (N, K) = {
    let v = read_vec();
    (v[0], v[1])
  };

  let mut v: Vec<(i64, i64)> = Vec::new();
  for _ in 0..N {
    let a = read_vec();
    v.push((a[0], a[1]));
  }
  let mut ans: u64 = std::u64::MAX;
  for &(_, up) in &v {
    for &(_, down) in &v {
      for &(left, _) in &v {
        for &(right, _) in &v {
          let mut tmp = 0;
          for &(i, j) in &v {
            count(i, j, left, right, up, down, &mut tmp);
          }
          if tmp == K {
            ans = min(ans, (left - right).abs() as u64 * (down - up).abs() as u64);
          }
        }
      }
    }
  }
  println!("{}", ans);
}

fn count(i: i64, j: i64, side1: i64, side2: i64, top1: i64, top2: i64, count: &mut u64) {
  let left = min(side1, side2);
  let right = max(side1, side2);
  let down = min(top1, top2);
  let top = max(top1, top2);

  if left <= i && i <= right && down <= j && j <= top {
    *count += 1;
  }

}