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 (1.42.0) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n: usize = read();
    let k: usize = read();
    let mut pv: Vec<(i64, i64)> = vec![];
    let mut minv = std::i64::MAX;

    for _ in 0..n {
        let x: i64 = read();
        let y: i64 = read();
        pv.push((x, y));
    }

    for u in 0..n {
        for d in 0..n {
            for r in 0..n {
                for l in 0..n {
                    let up = pv[u].1;
                    let down = pv[d].1;
                    let right = pv[r].0;
                    let left = pv[l].0;
                    if up <= down {
                        continue;
                    }
                    if right <= left {
                        continue;
                    }
                    let mut cnt = 0;
                    for i in 0..n {
                        let pi = pv[i];
                        if pi.1 <= up && pi.1 >= down && pi.0 <= right && pi.0 >= left {
                            cnt += 1;
                        }
                    }
                    if cnt >= k {
                        let sq = (up - down) * (right - left);
                        if sq < minv {
                            minv = sq;
                        }
                    }
                }
            }
        }
    }
    println!("{}", minv);
}