Problem:
Score : 600 points

Problem Statement

Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.

Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.

Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.

Constraints

All values in input are integers.

1 \leq N \leq 60

1 \leq K \leq N

-1000 \leq x_i , y_i \leq 1000

\left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)

1 \leq c_i \leq 100

Input

Input is given from Standard Input in the following format:

N K
x_1 y_1 c_1
\vdots
x_N y_N c_N

Output

Print the answer.

It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

4 3
-1 0 3
0 0 3
1 0 2
1 1 40

Sample Output 1

2.4

If we put the heat source at \left(-0.2, 0\right), the 1-st, 2-nd, and 3-rd pieces of meat will be ready to eat within 2.4 seconds. This is the optimal place to put the heat source.

Sample Input 2

10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16

Sample Output 2

7411.2252

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

const EPS: f64 = 1e-9;

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let k: usize = s.next().unwrap().parse().unwrap();
	let b: Vec<_> = (0..n).map(|_| (s.next().unwrap().parse::<f64>().unwrap(), s.next().unwrap().parse::<f64>().unwrap(), s.next().unwrap().parse::<f64>().unwrap())).collect();

	let mut l = 0.;
	let mut r = 1e8;
	for _ in 0..128 {
		let mid = (l + r) / 2.;

		if check(mid, k, &b) {
			r = mid;
		} else {
			l = mid;
		}
	}
	println!("{}", l);
}

fn check(t: f64, k: usize, b: &Vec<(f64, f64, f64)>) -> bool {
	let mut point = vec![];

	for i in 1..b.len() {
		for j in 0..i {
			let (x1, y1, r1) = (b[i].0, b[i].1, t / b[i].2);
			let (x2, y2, r2) = (b[j].0, b[j].1, t / b[j].2);
			let (dx, dy) = (x2 - x1, y2 - y1);
			let dd = dx * dx + dy * dy;
			if dd == 0. {
				continue;
			}
			let ad = (dd + r1 * r1 - r2 * r2) / 2.;
			let h = dd * r1 * r1 - ad * ad;
			if h <= 0. {
				continue;
			}
			let h = h.sqrt();
			let x3 = x1 + ad * dx / dd;
			let x4 = h * dy / dd;
			let y3 = y1 + ad * dy / dd;
			let y4 = h * dx / dd;
			point.push((x3 + x4, y3 - y4));
			point.push((x3 - x4, y3 + y4));
		}
	}

	for &(x, y, _) in b.iter() {
		point.push((x, y));
	}

	for (px, py) in point {
		let mut cnt = 0;
		for &(x, y, c) in b.iter() {
			let dx = x - px;
			let dy = y - py;
			let d = dx * dx + dy * dy;
			let r = t / c;
			if d < r * r + EPS {
				cnt += 1;
			}
		}
		if cnt >= k {
			return true;
		}
	}
	false
}