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: Runtime Error)

#2分探索
n,k = map(int, input().split( ))
m = [tuple(map(int, input().split( ))) for i in range(k)]

mx = 10**15
mn = 0
from math import sqrt
def solve(x1, y1, r1, x2, y2, r2):
    rr0 = (x2 - x1)**2 + (y2 - y1)**2
    xd = x2 - x1
    yd = y2 - y1
    rr1 = r1**2; rr2 = r2**2
    cv = (rr0 + rr1 - rr2)
    sv = sqrt(4*rr0*rr1 - cv**2)
    return (
        (x1 + (cv*xd - sv*yd)/(2.*rr0), y1 + (cv*yd + sv*xd)/(2.*rr0)),
        (x1 + (cv*xd + sv*yd)/(2.*rr0), y1 + (cv*yd - sv*xd)/(2.*rr0)),
    )


while mx-mn>10**(-8):
    md = (mx+mn)/2
    flag = True
    for i in range(n):
        for j in range(i):
            p = solve(m[i][0],m[i][1],md/m[i][2],m[j][0],m[j][1],md/m[j][2])
            for pi in p:
                flag = True
                for k in range(n):
                    if m[k][2]*((pi[0]-m[k][0])**2+(pi[1]-m[k][1])**2)**(1/2)>md:
                        flag = False
                        break
                
    if flag:
        mx = md
    else:
        mn = md
print(mx)