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)

use std::io::Read;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let m: usize = it.next().unwrap().parse().unwrap();
    if m == 1 {
        println!("0");
        return;
    }
    type T = (f64, f64, f64);
    let p: Vec<T> = (0..n).map(|_| {
        let x: f64 = it.next().unwrap().parse().unwrap();
        let y: f64 = it.next().unwrap().parse().unwrap();
        let c: f64 = it.next().unwrap().parse().unwrap();
        (x, y, c)
    }).collect();
    let eps = 1e-9;
    let distance = |a: (f64, f64), b: (f64, f64)| -> f64 {
        let dx = a.0 - b.0;
        let dy = a.1 - b.1;
        (dx * dx + dy * dy).sqrt()
    };
    let common = |a: T, b: T, t: f64| -> Option<((f64, f64), (f64, f64))> {
        let (p, q, r) = b;
        let (a, b, c) = a;
        let ra = t / c;
        let rb = t / r;
        if ra + rb - eps < distance((a, b), (p, q)) {
            return None;
        }
        let step = (a, b);
        let (x, y, r1, r2) = (p - a, q - b, ra, rb);
        let (a, b, c) = (2.0 * x, 2.0 * y, r2 * r2 - r1 * r1 - x * x - y * y);
        let (p, q, r) = (a * a + b * b, 2.0 * a * c, c * c - b * b * r1 * r1);
        let det = q * q - 4.0 * p * r;
        let xs = ((-q + det.sqrt()) / (2.0 * p), (-q - det.sqrt()) / (2.0 * p));
        let (p, q, r) = (a * a + b * b, 2.0 * b * c, c * c - a * a * r1 * r1);
        let det = q * q - 4.0 * p * r;
        let ys = ((-q - det.sqrt()) / (2.0 * p), (-q + det.sqrt()) / (2.0 * p));
        Some(((xs.0 + step.0, ys.0 + step.1), (xs.1 + step.0, ys.1 + step.1)))
    };
    let mut ng = 0f64;
    let mut ok = 3000f64 * 100f64;
    for _ in 0..100 {
        let mid = (ng + ok) * 0.5;
        let mut valid = false;
        for (i, &a) in p.iter().enumerate() {
            let mut cnt = 1;
            let a = (a.0, a.1);
            for (j, &b) in p.iter().enumerate() {
                if j != i && distance(a, (b.0, b.1)) * b.2 < mid + eps {
                    cnt += 1;
                }
            }
            if cnt >= m {
                valid = true;
            }
        }
        for (i, &a) in p.iter().enumerate() {
            for (j, &b) in p[..i].iter().enumerate() {
                if let Some((a, b)) = common(a, b, mid) {
                    for &t in [a, b].iter() {
                        let mut cnt = 2;
                        for (k, &(x, y, c)) in p.iter().enumerate() {
                            if k != i && k != j && c * distance((x, y), t) < mid + eps{
                                cnt += 1;
                            }
                        }
                        if cnt >= m {
                            valid = true;
                        }
                    }
                }
            }
        }
        if valid {
            ok = mid;
        } else {
            ng = mid;
        }
    }
    println!("{:.7}", ok);
}

fn main() {
    run();
}