#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <cassert>

using namespace std;
typedef long long LL;
template <typename F, typename S> std::ostream& operator<<(ostream& os, const pair<F,S>& p) {
    os << "(" << p.first << "," << p.second << ")"; return os; }
template <typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "["; for(const T& a: v){ os << a << ", "; } os << "]"; return os; }
template <typename K, typename V> ostream& operator<<(ostream& os, const map<K,V>& m) {
    os << "{"; for(const auto& p: m){ os << p.first <<":"<< p.second << ", "; } os << "}"; return os; }


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

double dist(double x1, double y1, double x2, double y2){
    double dy = y1 - y2;
    double dx = x1 - x2;
    return sqrt(dy*dy + dx*dx);
}

void circleCrossPoints(double x1, double y1, double r1, double x2, double y2, double r2, vector<pair<double, double>>& points){
    double d = sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
    if(d > r1 + r2) return;
    double rc = (d*d + r1*r1 - r2*r2) / (2*d);
    double rs = sqrt(r1*r1 - rc*rc);
    double dx = (x2 - x1) / d;
    double dy = (y2 - y1) / d;
    points.emplace_back(x1 + dx*rc - dy*rs, y1 + dy*rc + dx*rs);
    points.emplace_back(x1 + dx*rc + dy*rs, y1 + dy*rc - dx*rs);
}

void addPoints(Niku& a, Niku& b, double t, vector<pair<double, double>>& points){
    double r1 = t / a.c;
    double r2 = t / b.c;
    circleCrossPoints(a.x, a.y, r1, b.x, b.y, r2, points);
}


bool acceptable(pair<double, double>& p, LL k, vector<Niku>& niku, double t){
    LL cnt = 0;
    for(Niku a: niku){
        double r = dist(p.first, p.second, a.x, a.y);
        if(r < t / a.c + 1.0e-9) cnt++;
    }
    return cnt >= k;
}


bool acceptable(LL n, LL k, vector<Niku>& niku, double t){
    vector<pair<double, double>> points;
    for(LL i = 0; i < n; ++i){
        points.emplace_back(niku[i].x, niku[i].y);
        for(LL j = i+1; j < n; ++j){
            addPoints(niku[i], niku[j], t, points);
        }
    }
    for(auto p: points){
        if(acceptable(p, k, niku, t)) return true;
    }
    return false;
}

int main(){
    LL n, k;
    cin >> n >> k;
    vector<Niku> niku(n);
    for(LL i = 0; i < n; ++i){
        cin >> niku[i].x >> niku[i].y >> niku[i].c;
    }
    double right = 10000000;
    double left = 0;
    for(LL e = 0; e < 50; ++e){
        double middle = (left + right) / 2;
        if(acceptable(n, k, niku, middle)){
            right = middle;
        }else{
            left = middle;
        }
    }
    cout << setprecision(13) << left << 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 <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <cassert>

using namespace std;
typedef long long LL;
template <typename F, typename S> std::ostream& operator<<(ostream& os, const pair<F,S>& p) {
    os << "(" << p.first << "," << p.second << ")"; return os; }
template <typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "["; for(const T& a: v){ os << a << ", "; } os << "]"; return os; }
template <typename K, typename V> ostream& operator<<(ostream& os, const map<K,V>& m) {
    os << "{"; for(const auto& p: m){ os << p.first <<":"<< p.second << ", "; } os << "}"; return os; }


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

double dist(double x1, double y1, double x2, double y2){
    double dy = y1 - y2;
    double dx = x1 - x2;
    return sqrt(dy*dy + dx*dx);
}

void circleCrossPoints(double x1, double y1, double r1, double x2, double y2, double r2, vector<pair<double, double>>& points){
    double d = sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
    if(d > r1 + r2) return;
    double rc = (d*d + r1*r1 - r2*r2) / (2*d);
    double rs = sqrt(r1*r1 - rc*rc);
    double dx = (x2 - x1) / d;
    double dy = (y2 - y1) / d;
    points.emplace_back(x1 + dx*rc - dy*rs, y1 + dy*rc + dx*rs);
    points.emplace_back(x1 + dx*rc + dy*rs, y1 + dy*rc - dx*rs);
}

void addPoints(Niku& a, Niku& b, double t, vector<pair<double, double>>& points){
    double r1 = t / a.c;
    double r2 = t / b.c;
    circleCrossPoints(a.x, a.y, r1, b.x, b.y, r2, points);
}


bool acceptable(pair<double, double>& p, LL k, vector<Niku>& niku, double t){
    LL cnt = 0;
    for(Niku a: niku){
        double r = dist(p.first, p.second, a.x, a.y);
        if(r < t / a.c + 1.0e-9) cnt++;
    }
    return cnt >= k;
}


bool acceptable(LL n, LL k, vector<Niku>& niku, double t){
    vector<pair<double, double>> points;
    for(LL i = 0; i < n; ++i){
        points.emplace_back(niku[i].x, niku[i].y);
        for(LL j = i+1; j < n; ++j){
            addPoints(niku[i], niku[j], t, points);
        }
    }
    for(auto p: points){
        if(acceptable(p, k, niku, t)) return true;
    }
    return false;
}

int main(){
    LL n, k;
    cin >> n >> k;
    vector<Niku> niku(n);
    for(LL i = 0; i < n; ++i){
        cin >> niku[i].x >> niku[i].y >> niku[i].c;
    }
    double right = 10000000;
    double left = 0;
    for(LL e = 0; e < 50; ++e){
        double middle = (left + right) / 2;
        if(acceptable(n, k, niku, middle)){
            right = middle;
        }else{
            left = middle;
        }
    }
    cout << setprecision(13) << left << endl;
    return 0;
}


```