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: Wrong Answer)

use proconio::input;
#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars};
#[allow(unused_imports)]
use std::cmp::{min, max};

fn main() {
	input! {
		n: usize,
		k: usize,
		sxy: [(isize, isize); n],
	}

	let mut x = vec![0usize; n];
	let mut y = vec![0usize; n];
	for i in 0..n {
		x[i] = (sxy[i].0 + 1000000000) as usize;
		y[i] = (sxy[i].1 + 1000000000) as usize;
	}

	x.sort();
	y.sort();

	let mut ans = (x.last().unwrap() - x[0])*(y.last().unwrap() - y[0]);

	for xi in 0..n-1 {
		for xj in xi+1..n {
			for yi in 0..n-1 {
				for yj in yi+1..n {
					let mut count = 0;
					for i in 0..n {
						if x[xi] <= x[i] && x[i] <= x[xj] && y[yi] <= y[i] && y[i] <= y[yj] {
							count += 1;
						}
					}
					if count >= k {
						ans = min(ans, (x[xj]-x[xi])*(y[yj]-y[yi]));
					}
				}
			}
		}
	}
	println!("{}", ans);
}