#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
void tostr(ll x,string& res){while(x)res+=('0'+(x%10)),x/=10; reverse(ALL(res)); return;}
template<class T> inline bool chmax(T& a,T b){ if(a<b){a=b;return 1;}return 0; }
template<class T> inline bool chmin(T& a,T b){ if(a>b){a=b;return 1;}return 0; }
//end

using  point=complex<double>;
struct circle{point p; double r;};
bool check(circle c1,circle c2){
   if(c1.r<c2.r)swap(c1,c2);
   double d=abs(c1.p-c2.p); if(c1.r+c2.r<d)return 0;
   if(abs(c1.r+c2.r-d)<eps||c1.r-c2.r<d||abs(c1.r-c2.r-d)<eps)return 1;
   return 0;
}
vector<point> intersect(circle c1,circle c2){
   if(!check(c1,c2))return {};
   double d=abs(c1.p-c2.p);
   double a=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
   double t=atan2(c2.p.imag()-c1.p.imag(),c2.p.real()-c1.p.real());
   auto r1=polar(c1.r,t+a)+c1.p,r2=polar(c1.r,t-a)+c1.p;
   return {r1,r2};
}

int n,k; double x[61],y[61],c[61];
circle cs[61];
int cnt(point p){
   int res=0;
   rep(i,0,n){if(abs(cs[i].p-p)<=cs[i].r+eps)res++;}
   return res;
}

int main(){
   scanf("%d%d",&n,&k);
   rep(i,0,n)scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);
   double lb=0,rb=1e6;
   rep(_,0,100){
      double mid=(lb+rb)/2.0;
      rep(i,0,n)cs[i].p=point(x[i],y[i]),cs[i].r=mid/c[i];
      auto f=[&](){ 
         vector<point> cand;
         rep(i,0,n){
            cand.push_back(cs[i].p);
            rep(j,i+1,n){
               auto v=intersect(cs[i],cs[j]);
               for(auto p:v)cand.push_back(p);
            }
         }
         for(point p:cand)if(cnt(p)>=k)return 1;
         return 0;
      };
      if(f())rb=mid; else lb=mid;
   }
   printf("%.12f\n",rb);
   return 0;
}
### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
void tostr(ll x,string& res){while(x)res+=('0'+(x%10)),x/=10; reverse(ALL(res)); return;}
template<class T> inline bool chmax(T& a,T b){ if(a<b){a=b;return 1;}return 0; }
template<class T> inline bool chmin(T& a,T b){ if(a>b){a=b;return 1;}return 0; }
//end

using  point=complex<double>;
struct circle{point p; double r;};
bool check(circle c1,circle c2){
   if(c1.r<c2.r)swap(c1,c2);
   double d=abs(c1.p-c2.p); if(c1.r+c2.r<d)return 0;
   if(abs(c1.r+c2.r-d)<eps||c1.r-c2.r<d||abs(c1.r-c2.r-d)<eps)return 1;
   return 0;
}
vector<point> intersect(circle c1,circle c2){
   if(!check(c1,c2))return {};
   double d=abs(c1.p-c2.p);
   double a=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
   double t=atan2(c2.p.imag()-c1.p.imag(),c2.p.real()-c1.p.real());
   auto r1=polar(c1.r,t+a)+c1.p,r2=polar(c1.r,t-a)+c1.p;
   return {r1,r2};
}

int n,k; double x[61],y[61],c[61];
circle cs[61];
int cnt(point p){
   int res=0;
   rep(i,0,n){if(abs(cs[i].p-p)<=cs[i].r+eps)res++;}
   return res;
}

int main(){
   scanf("%d%d",&n,&k);
   rep(i,0,n)scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);
   double lb=0,rb=1e6;
   rep(_,0,100){
      double mid=(lb+rb)/2.0;
      rep(i,0,n)cs[i].p=point(x[i],y[i]),cs[i].r=mid/c[i];
      auto f=[&](){ 
         vector<point> cand;
         rep(i,0,n){
            cand.push_back(cs[i].p);
            rep(j,i+1,n){
               auto v=intersect(cs[i],cs[j]);
               for(auto p:v)cand.push_back(p);
            }
         }
         for(point p:cand)if(cnt(p)>=k)return 1;
         return 0;
      };
      if(f())rb=mid; else lb=mid;
   }
   printf("%.12f\n",rb);
   return 0;
}
```