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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;

// template {{{  0 
// using {{{ 1
using ll = long long int;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
// }}} 1
// definition {{{ 1
// scaning {{{ 2
#define Scd(x) scanf("%d", &x)
#define Scd2(x,y) scanf("%d%d", &x, &y)
#define Scd3(x,y,z) scanf("%d%d%d", &x, &y, &z)
#define Scll(x) scanf("%lld", &x)
#define Scll2(x,y) scanf("%lld%lld", &x, &y)
#define Scll3(x,y,z) scanf("%lld%lld%lld", &x, &y, &z)
#define Scc(c) scanf("%c", &c);
#define Scs(s) scanf("%s", s);
#define Scstr(s) scanf("%s", &s);
// }}} 2
// constants {{{ 2
#define EPS (1e-7)
#define INF (2e9)
#define PI (acos(-1))
// }}} 2
// systems {{{ 2
#define Repe(x,y,z) for(ll x = z; x < y; x++)
#define Rep(x,y) Repe(x,y,0)
#define RRepe(x,y,z) for(ll x = y-z-1; x >= 0; x--)
#define RRep(x,y) RRepe(x,y,0)
// }}} 2
// output {{{ 2
#define YesNo(a) (a)?printf("Yes\n"):printf("No\n")
#define YESNO(a) (a)?printf("YES\n"):printf("NO\n")
// }}} 2
// }}} 1
// input {{{ 1
// }}} 1
// }}} 0

struct nick{
    double x,y,c;
};

int main() {

    int N,K;
    Scd2(N,K);

    auto dis = []( double X, double Y, nick nik ){
        double a = (X-nik.x); double b = (Y-nik.y);
        return nik.c*sqrt(a*a+b*b);
    };

    vector<nick> n(N);
    int x,y,c;
    Rep(i,N){
        Scd3(x,y,c);
        n[i] = {(double)x,(double)y,(double)c};
    }

    double tmp;
    double ans = 1e60;
    if( K == 1 ) ans = 0;
    else{
        Rep(i,N) Repe(j,N,i){
            tmp = 0;
            double X = (n[i].c*n[i].x + n[j].c*n[j].x) / (n[j].c+n[i].c);
            double Y = (n[i].c*n[i].y + n[j].c*n[j].y) / (n[j].c+n[i].c);
            priority_queue<double> q;
            Rep(k,N){
                q.push( dis(X,Y,n[k]) );
            }
            Rep(k,N-K) q.pop();
            tmp = q.top();
            Rep(k,K-1) q.pop();
            ans = min(tmp,ans);
        }
    }

    printf ("%.9lf\n", ans );

    return 0;
}