#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> p_ll;

template<class T>
void debug(T itr1, T itr2) { auto now = itr1; while(now<itr2) { cout << *now << " "; now++; } cout << endl; }
#define repr(i,from,to) for (int i=(int)from; i<(int)to; i++)
#define all(vec) vec.begin(), vec.end()
#define rep(i,N) repr(i,0,N)
#define per(i,N) for (int i=(int)N-1; i>=0; i--)

const ll MOD = pow(10,9)+7;
const ll LLINF = pow(2,61)-1;
const int INF = pow(2,30)-1;

vector<ll> fac;
void c_fac(int x=pow(10,6)+10) { fac.resize(x,true); rep(i,x) fac[i] = i ? (fac[i-1]*i)%MOD : 1; }
ll inv(ll a, ll m=MOD) { ll b = m, x = 1, y = 0; while (b!=0) { int d = a/b; a -= b*d; swap(a,b); x -= y*d; swap(x,y); } return (x+m)%m; }
ll nck(ll n, ll k) { return fac[n]*inv(fac[k]*fac[n-k]%MOD)%MOD; }
ll gcd(ll a, ll b) { if (a<b) swap(a,b); return b==0 ? a : gcd(b, a%b); }
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; }

struct niku { double x; double y; double c; };

int main() {
  int N, K; cin >> N >> K;
  niku n[N];
  rep(i,N) {
    double x, y, c; cin >> x >> y >> c;
    n[i] = {x,y,c};
  }
  double l = 0, r = 5000*100;
  while(r-l>=1e-8) {
    double mid = (l+r)/2;
    bool ok = false;
    rep(i,N) repr(j,i,N) {
      if (i==j) {
        int count = 0;
        rep(k,N) if (pow(n[k].x-n[i].x,2)+pow(n[k].y-n[i].y,2)<=pow(mid/n[k].c,2)) count++;
        if (count>=K) ok = true;
      }
      else {
        double ri = mid/n[i].c, rj = mid/n[j].c;
        if (pow(n[i].x-n[j].x,2)+pow(n[i].y-n[j].y,2)>pow(ri+rj,2)) continue;
        if (pow(n[i].x-n[j].x,2)+pow(n[i].y-n[j].y,2)<pow(ri-rj,2)) continue;
        double vx = n[i].x-n[j].x, vy = n[i].y-n[j].y;
        double vl = sqrt(vx*vx + vy*vy);
        vx /= vl; vy /= vl;
        double dl = (ri*ri-rj*rj+vl*vl) / (2*vl);
        double rx = n[i].x - vx*dl, ry = n[i].y - vy*dl;
        double rx1 = rx - vy * sqrt(ri*ri-dl*dl), ry1 = ry + vx * sqrt(ri*ri-dl*dl);
        double rx2 = rx + vy * sqrt(ri*ri-dl*dl), ry2 = ry - vx * sqrt(ri*ri-dl*dl);
        int count1 = 0, count2 = 0;
        rep(k,N) if (pow(n[k].x-rx1,2)+pow(n[k].y-ry1,2)<=pow(mid/n[k].c,2)+1e-6) count1++;
        rep(k,N) if (pow(n[k].x-rx2,2)+pow(n[k].y-ry2,2)<=pow(mid/n[k].c,2)+1e-6) count2++;
        if (count1>=K||count2>=K) ok = true;
        // cout << mid << " " << i << " " << j << " " << rx1 << " " << ry1 << " " << count1 << endl;
        // cout << mid << " " << i << " " << j << " " << rx2 << " " << ry2 << " " << count2 << endl;
      }
    }
    // cout << l << " " << mid << " " << r << endl;
    if (ok) r = mid;
    else l = mid;
  }

  double result = l;
  cout << setprecision(10) << result << endl;
  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
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> p_ll;

template<class T>
void debug(T itr1, T itr2) { auto now = itr1; while(now<itr2) { cout << *now << " "; now++; } cout << endl; }
#define repr(i,from,to) for (int i=(int)from; i<(int)to; i++)
#define all(vec) vec.begin(), vec.end()
#define rep(i,N) repr(i,0,N)
#define per(i,N) for (int i=(int)N-1; i>=0; i--)

const ll MOD = pow(10,9)+7;
const ll LLINF = pow(2,61)-1;
const int INF = pow(2,30)-1;

vector<ll> fac;
void c_fac(int x=pow(10,6)+10) { fac.resize(x,true); rep(i,x) fac[i] = i ? (fac[i-1]*i)%MOD : 1; }
ll inv(ll a, ll m=MOD) { ll b = m, x = 1, y = 0; while (b!=0) { int d = a/b; a -= b*d; swap(a,b); x -= y*d; swap(x,y); } return (x+m)%m; }
ll nck(ll n, ll k) { return fac[n]*inv(fac[k]*fac[n-k]%MOD)%MOD; }
ll gcd(ll a, ll b) { if (a<b) swap(a,b); return b==0 ? a : gcd(b, a%b); }
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; }

struct niku { double x; double y; double c; };

int main() {
  int N, K; cin >> N >> K;
  niku n[N];
  rep(i,N) {
    double x, y, c; cin >> x >> y >> c;
    n[i] = {x,y,c};
  }
  double l = 0, r = 5000*100;
  while(r-l>=1e-8) {
    double mid = (l+r)/2;
    bool ok = false;
    rep(i,N) repr(j,i,N) {
      if (i==j) {
        int count = 0;
        rep(k,N) if (pow(n[k].x-n[i].x,2)+pow(n[k].y-n[i].y,2)<=pow(mid/n[k].c,2)) count++;
        if (count>=K) ok = true;
      }
      else {
        double ri = mid/n[i].c, rj = mid/n[j].c;
        if (pow(n[i].x-n[j].x,2)+pow(n[i].y-n[j].y,2)>pow(ri+rj,2)) continue;
        if (pow(n[i].x-n[j].x,2)+pow(n[i].y-n[j].y,2)<pow(ri-rj,2)) continue;
        double vx = n[i].x-n[j].x, vy = n[i].y-n[j].y;
        double vl = sqrt(vx*vx + vy*vy);
        vx /= vl; vy /= vl;
        double dl = (ri*ri-rj*rj+vl*vl) / (2*vl);
        double rx = n[i].x - vx*dl, ry = n[i].y - vy*dl;
        double rx1 = rx - vy * sqrt(ri*ri-dl*dl), ry1 = ry + vx * sqrt(ri*ri-dl*dl);
        double rx2 = rx + vy * sqrt(ri*ri-dl*dl), ry2 = ry - vx * sqrt(ri*ri-dl*dl);
        int count1 = 0, count2 = 0;
        rep(k,N) if (pow(n[k].x-rx1,2)+pow(n[k].y-ry1,2)<=pow(mid/n[k].c,2)+1e-6) count1++;
        rep(k,N) if (pow(n[k].x-rx2,2)+pow(n[k].y-ry2,2)<=pow(mid/n[k].c,2)+1e-6) count2++;
        if (count1>=K||count2>=K) ok = true;
        // cout << mid << " " << i << " " << j << " " << rx1 << " " << ry1 << " " << count1 << endl;
        // cout << mid << " " << i << " " << j << " " << rx2 << " " << ry2 << " " << count2 << endl;
      }
    }
    // cout << l << " " << mid << " " << r << endl;
    if (ok) r = mid;
    else l = mid;
  }

  double result = l;
  cout << setprecision(10) << result << endl;
  return 0;
}
```