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: Wrong Answer)

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 mut x: Vec<Vec<i64>> = (0..n)
        .map(|_| {
            let x = input();
            let y = input();
            vec![x, y]
        })
        .collect();
    x.sort_by_key(|v| v[0] + v[1]);

    let mut ans = i64::max_value();
    for p in x.windows(k) {
        let mut x_min = i64::max_value();
        let mut y_min = i64::max_value();
        let mut x_max = i64::min_value();
        let mut y_max = i64::min_value();
        for q in p {
            x_min = std::cmp::min(x_min, q[0]);
            y_min = std::cmp::min(y_min, q[1]);
            x_max = std::cmp::max(x_max, q[0]);
            y_max = std::cmp::max(y_max, q[1]);
        }
        ans = std::cmp::min(ans, (x_max - x_min) * (y_max - y_min));
    }

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