#include<iostream>
#include<iomanip>
#include<algorithm>
#include<numeric>
#include<cmath>
#include<queue>
#include<tuple>

int const MAXN = 60;
double const eps = 5e-7;

int x[MAXN], y[MAXN], c[MAXN], c2[MAXN], ix[MAXN], N, K;
double d2[MAXN];

double min_possible(double x1, double y1, double r1) {
  for(int i = 0; i < N; ++i) {
    double dx = x[i] - x1, dy = y[i] - y1;
    dx = std::max(std::fabs(dx) - r1, 0.0); // min of abs in [dx - r1, dx + r1]
    dy = std::max(std::fabs(dy) - r1, 0.0); // similar
    //d2[i] = (dx*dx + dy*dy)*c2[i];
    d2[i] = std::hypot(dx, dy)*c[i];
  }
  std::nth_element(d2, d2+K-1, d2+N);
  //return std::sqrt(d2[K-1]);
  return d2[K-1];
}

double calc(double xx, double yy) {
  for(int i = 0; i < N; ++i) {
    double dx = x[i] - xx, dy = y[i] - yy;
    //d2[i] = (dx*dx + dy*dy)*c2[i];
    d2[i] = hypot(dx, dy)*c[i];
  }
  std::nth_element(d2, d2+K-1, d2+N);
  //return std::sqrt(d2[K-1]);
  return d2[K-1];
}

int main() {
  std::cin >> N >> K;
  for(int i = 0; i < N; ++i) {
    std::cin >> x[i] >> y[i] >> c[i];
    c2[i] = c[i]*c[i];
  }
  std::queue<std::tuple<double, double, double>> q;
  q.emplace(0.0, 0.0, 1000.0);
  double ans = 1e9;
  while(q.size() > 0) {
    double xx, yy, rr;
    std::tie(xx, yy, rr) = q.front(); q.pop();
    double d = min_possible(xx, yy, rr);
    if((ans-d) <= eps*std::max(d, 1.0)) continue;
    ans = std::min(ans, calc(xx, yy));
    for(double rx:{-0.5, 0.5}) for(double ry:{-0.5, 0.5}) {
      q.emplace(xx+rx*rr, yy+ry*rr, rr*0.5);
    }
  }
  std::cout << std::setprecision(10) << std::fixed << ans << std::endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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<iostream>
#include<iomanip>
#include<algorithm>
#include<numeric>
#include<cmath>
#include<queue>
#include<tuple>

int const MAXN = 60;
double const eps = 5e-7;

int x[MAXN], y[MAXN], c[MAXN], c2[MAXN], ix[MAXN], N, K;
double d2[MAXN];

double min_possible(double x1, double y1, double r1) {
  for(int i = 0; i < N; ++i) {
    double dx = x[i] - x1, dy = y[i] - y1;
    dx = std::max(std::fabs(dx) - r1, 0.0); // min of abs in [dx - r1, dx + r1]
    dy = std::max(std::fabs(dy) - r1, 0.0); // similar
    //d2[i] = (dx*dx + dy*dy)*c2[i];
    d2[i] = std::hypot(dx, dy)*c[i];
  }
  std::nth_element(d2, d2+K-1, d2+N);
  //return std::sqrt(d2[K-1]);
  return d2[K-1];
}

double calc(double xx, double yy) {
  for(int i = 0; i < N; ++i) {
    double dx = x[i] - xx, dy = y[i] - yy;
    //d2[i] = (dx*dx + dy*dy)*c2[i];
    d2[i] = hypot(dx, dy)*c[i];
  }
  std::nth_element(d2, d2+K-1, d2+N);
  //return std::sqrt(d2[K-1]);
  return d2[K-1];
}

int main() {
  std::cin >> N >> K;
  for(int i = 0; i < N; ++i) {
    std::cin >> x[i] >> y[i] >> c[i];
    c2[i] = c[i]*c[i];
  }
  std::queue<std::tuple<double, double, double>> q;
  q.emplace(0.0, 0.0, 1000.0);
  double ans = 1e9;
  while(q.size() > 0) {
    double xx, yy, rr;
    std::tie(xx, yy, rr) = q.front(); q.pop();
    double d = min_possible(xx, yy, rr);
    if((ans-d) <= eps*std::max(d, 1.0)) continue;
    ans = std::min(ans, calc(xx, yy));
    for(double rx:{-0.5, 0.5}) for(double ry:{-0.5, 0.5}) {
      q.emplace(xx+rx*rr, yy+ry*rr, rr*0.5);
    }
  }
  std::cout << std::setprecision(10) << std::fixed << ans << std::endl;
  return 0;
}

```