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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<sys/time.h>

unsigned long long seed;

unsigned long long xorshift(){
  seed=seed^(seed<<13);
  seed=seed^(seed>>7);
  seed=seed^(seed<<17);
  return seed;
}

double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}

typedef struct{
double val;
int id;
}sd;

int sdsortfnc(const void *a,const void *b){
if(((sd*)a)->val < ((sd*)b)->val){return -1;}
if(((sd*)a)->val > ((sd*)b)->val){return 1;}
return 0;
}

int n,k;
double ix[1024],iy[1024],ic[1024];
sd mem[1024];

int eval(double a,double b,double *res){
  int i;
  for(i=0;i<n;i++){
    mem[i].val=ic[i]*sqrt(dbzt(a,ix[i])*dbzt(a,ix[i])+dbzt(b,iy[i])*dbzt(b,iy[i]));
    mem[i].id=i;
  }
  qsort(mem,n,sizeof(mem[0]),sdsortfnc);
  (*res)=mem[k-1].val;
  return mem[k-1].id;
}

int main(){
  struct timeval tv;
  gettimeofday(&tv, NULL);
  seed=(tv.tv_sec*1000000+tv.tv_usec);
  int i,t,p;
  scanf("%d%d",&n,&k);
  double rx=0.0,ry=0.0,res,fres=1.0e18,temp;
  for(i=0;i<n;i++){
    scanf("%lf%lf%lf",&ix[i],&iy[i],&ic[i]);
  }
  for(t=0;t<50;t++){
    rx=(xorshift()%20000001);
    rx/=10000.0;
    rx-=1000.0;
    ry=(xorshift()%20000001);
    ry/=10000.0;
    ry-=1000.0;
    temp=0.01;
    for(i=0;i<10000;i++){
      p=eval(rx,ry,&res);
      fres=dbmin(res,fres);
      rx+=(ix[p]-rx)*temp;
      ry+=(iy[p]-ry)*temp;
      temp*=0.998;
    }
  }
  printf("%.12lf\n",fres);
}