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::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let k: usize = input();
    let xy: Vec<(i64, i64)> = (0..n)
        .map(|_| {
            let x = input();
            let y = input();
            (x, y)
        })
        .collect();
    let (mut qx, mut qy): (Vec<i64>, Vec<i64>) = xy.iter().cloned().unzip::<i64, i64, _, _>();
    qx.sort();
    qy.sort();

    let mut ans = i64::max_value();
    for xi in 0..n {
        for xj in xi + 1..n {
            for yi in 0..n {
                for yj in yi + 1..n {
                    let mut cnt = 0;
                    for m in 0..n {
                        let (px, py) = xy[m];
                        if qx[xi] <= px && px <= qx[xj] && qy[yi] <= py && py <= qy[yj] {
                            cnt += 1;
                        }
                    }
                    if cnt >= k {
                        ans = std::cmp::min(ans, (qx[xi] - qx[xj]).abs() * (qy[yi] - qy[yj]).abs());
                    }
                }
            }
        }
    }

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