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;
use std::cmp;

fn gets() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    return buf
}

fn main() {
    let line = gets();
    let mut it = line.trim().split_whitespace().map(|n| n.parse::<usize>().unwrap());
    let (n, k) = (it.next().unwrap(), it.next().unwrap());
    let mut px = vec![0; n];
    let mut py = vec![0; n];
    let mut pxx = vec![0; n];
    let mut pyy = vec![0; n];
    for i in 0..n {
        let line = gets();
        let mut it = line.trim().split_whitespace().map(|n| n.parse::<i64>().unwrap());
        let (x, y) = (it.next().unwrap(), it.next().unwrap());
        px[i] = x;
        py[i] = y;
        pxx[i] = x;
        pyy[i] = y;
    }
    pxx.sort();
    pyy.sort();

    let mut ans = (pxx[n-1] - pxx[0]) * (pyy[n-1] - pyy[0]);
    for xi in 0..n {
        for xj in (xi + 1)..n {
            for yi in 0..n {
                for yj in (yi + 1)..n {
                    let lx = pxx[xi];
                    let rx = pxx[xj];
                    let by = pyy[yi];
                    let uy = pyy[yj];
                    let mut num = 0;
                    for i in 0..n {
                        if lx <= px[i] && rx >= px[i] && by <= py[i] && uy >= py[i] {
                            num += 1;
                        }
                    }
                    if num >= k {
                        let area = (rx - lx) * (uy - by);
                        ans = cmp::min(ans, area);
                    }
                } 
            }
        }
    }
    println!("{}", ans);
}