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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!/usr/bin/env python3
# %%
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import itertools

# %%
N, K = map(int, readline().split())
m = map(int, read().split())
XYC = zip(m, m, m)
X, Y, C = zip(*XYC)
X = [complex(x, y) for x, y in zip(X, Y)]

# %%


def solve_triple(x0, c0, x1, c1, x2, c2):
    x = (x0 + x1 + x2) / 3
    coef = 1
    for _ in range(200):
        d0 = c0 * abs(x0 - x)
        d1 = c1 * abs(x1 - x)
        d2 = c2 * abs(x2 - x)
        if d0 >= d1 and d0 >= d2:
            dx = x0 - x
        elif d1 >= d0 and d1 >= d2:
            dx = x1 - x
        elif d2 >= d0 and d2 >= d1:
            dx = x2 - x
        x += coef * dx
        coef *= 0.9
    return x


# %%
def gen_candidates():
    for i, j, k in itertools.combinations(range(N), 3):
        yield solve_triple(X[i], C[i], X[j], C[j], X[k], C[k])
    for i, j in itertools.combinations(range(N), 2):
        yield (C[i] * X[i] + C[j] * X[j]) / (C[i] + C[j])
    for i in range(N):
        yield X[i]


def solve(x0):
    dist = sorted(c * abs(x - x0) for x, c in zip(X, C))
    return dist[K - 1]


# %%
answer = min(map(solve, gen_candidates()))
print(answer)

# %%