Problem: 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}.

Examples

Input

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


Output

2.4


Input

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


Output

7411.2252

Solution (in PYTHON3):

Solution:
#2分探索
#md/cで交点が定まらない時の対案
n,k = map(int, input().split( ))
m = [tuple(map(int, input().split( ))) for i in range(n)]
#そもそも設定はn枚すべて焼く必要はない
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)
    if max(r1,r2,sqrt(rr0))*2>r1+r2+sqrt(rr0):
        return (x1,y1),(x2,y2)
    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)),
    )

#flag で breakしてたのが煩雑すぎて追えていなかったらしい
while mx-mn>10**(-6) and mx-mn>10**(-7)*mx:#mx大きい時に安易に設定すると到達しない？
    md = (mx+mn)/2
    cnt = 0
    for i in range(n):##中心選ぶケースが抜けてる
        pi0,pi1 = m[i][0],m[i][1]
        tmp = 0
        for l in range(n):#kはすでに使っている変数
            if m[l][2]*sqrt(((pi0-m[l][0])**2) + ((pi1-m[l][1])**2))<=md+0.000000001:
                tmp += 1
        if cnt<tmp:
            cnt = tmp

    for i in range(n):
        for j in range(i+1,n):##range(i)だとi=0で止まって何もしない

            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:
                tmp = 0
                for l in range(n):#kはすでに使っている変数
                    if (((pi[0]-m[l][0])**2) + ((pi[1]-m[l][1])**2))**(1/2)<=md/m[l][2]+0.000000001:##ここで誤差認めなかったせいらしい
                        tmp += 1
                if cnt<tmp:
                    cnt = tmp

    if cnt>=k:
        mx = md
    else:
        mn = md
print(mx)