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:
import cmath,sys
input = sys.stdin.readline


"""
cdist(z1,z2) : z1とz2の距離
cInDiv(z1,z2) : z1とz2をp:qに内分する点
cOutDiv(z1,z2) : z1とz2をp:qに外分する点
"""

# z.real,z.imag
EPS = 0.000000001

def cdist(z1,z2=0):
    return abs(z1-z2)

def cInDiv(z1,z2,p,q):
    return (q*z1+p*z2)/(p+q)

def cOutDiv(z1,z2,p,q):
    return (-q*z1+p*z2)/(p-q)

"""
cInCircle(z,c,r) : 点zは中心c半径rの円の中にあるか(0:円周上,-1:円内,1:円外)
cIntersectionPnt(z1,z2,r1,r2) : 中心z1半径r1,中心z2半径r2の円の交点の数、座標1,座標2
"""

def cInCircle(z,c,r):
    d = abs(z-c)
    if abs(r-d) < EPS:
        return 0
    if r-d > 0:
        return -1
    return 1

def cIntersectionPnt(z1,z2,r1,r2):
    d = abs(z1-z2)

    if d < EPS:
        if abs(r1-r2) < EPS:
            return float("inf"),-1,-1
        else:
            return 0,-1,-1
    
    
    if abs(d - (r1+r2)) < EPS:
        diff = (z2-z1)/d
        return 1,z1+diff*r1,-1
    
    if abs(d - (r1-r2)) < EPS:
        diff = (z2-z1)/d
        return 1,z1+diff*r1,-1
    
    if abs(d - (r2-r1)) < EPS:
        diff = (z1-z2)/d
        return 1,z2+diff*r2,-1

    if d > r1+r2:
        return 0,-1,-1
    if d < abs(r1-r2):
        return 0,-1,-1


    rc = (r1*r1+d*d-r2*r2)/(2*d)
    rs = cmath.sqrt(r1*r1-rc*rc)
    diff = (z2-z1)/d
    return 2,z1+diff*rc+diff*1j*rs,z1+diff*rc+diff*(-1j)*rs


n,k = map(int,input().split())
xy = []
c = []
for i in range(n):
    xx,yy,cc = map(int,input().split())
    xy.append(xx+yy*1j)
    c.append(cc)


def judge(t):
    points = []
    for i in range(n):
        for j in range(i+1,n):
            flag,pt1,pt2 = cIntersectionPnt(xy[i],xy[j],t/c[i],t/c[j])
            if flag == 1:
                points.append(pt1)
            elif flag == 2:
                points.append(pt1)
                points.append(pt2)
    for pc in xy:
        points.append(pc)
    for z in points:
        cnt = 0
        for i in range(n):
            flag = cInCircle(z,xy[i],t/c[i])
            if flag <= 0:
                cnt += 1
        if cnt >= k:
            return True
    return False


ok = 10**10
ng = 0
while abs(ok-ng) > EPS:
    mid = (ok+ng)/2
    if judge(mid):
        ok = mid
    else:
        ng = mid

print(ok)