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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int64_t n, k;
int64_t x[60], y[60], c[60];
double t;

using pt = pair<double, double>;

pt operator+(pt lhs, pt rhs){ return {lhs.first + rhs.first, lhs.second + rhs.second}; }
pt operator-(pt lhs, pt rhs){ return {lhs.first - rhs.first, lhs.second - rhs.second}; }
pt operator*(pt lhs, double rhs){ return {lhs.first * rhs, lhs.second * rhs}; }

double dist(pt p1, pt p2){
    return sqrt( pow(p1.first-p2.first, 2.0) + pow(p1.second-p2.second, 2.0) );
}

pair<bool, pair<pt, pt>> getpt(int i, int j){
    long double r1 = t / c[i];
    long double r2 = t / c[j];
    long double d = dist({x[i], y[i]}, {x[j], y[j]});
    if(r1 + r2 < d) return {false, {}};
    if(abs(r1 - r2) > d) return {false, {}};
    long double d1 = (d * d + r1 * r1 - r2 * r2) / (2.0 * d);
    long double d2 = (d * d - r1 * r1 + r2 * r2) / (2.0 * d);
    long double xx = r1 * r1 - d1 * d1;
    if(xx < 0) return {false, {}};
    pt rr = { (x[i] - x[j]) / d,  (y[i] - y[j]) / d };
    pt v  = { (y[i] - y[j]) / d, -(x[i] - x[j]) / d };
    pt ret1 = pt{ x[j], y[j] } + rr * d2;
    pt ret2 = v * sqrt(xx);
    return {true, {ret1, ret2}};
}

bool check(pt point){
    int cnt = 0;
    for(int i=0;i<n;++i){
        if(c[i] * dist({x[i], y[i]}, point) <= t){
            ++cnt;
            if(cnt >= k) return true;
        }
    }
    return false;
}

int main(){
    cout.precision(12);
    cout.setf(ios_base::fixed, ios_base::floatfield);

    cin >> n >> k;
    for(int i=0;i<n;++i){
        cin >> x[i] >> y[i] >> c[i];
    }
    pt dx = { 1e-8, 0.0 }, dy = { 0.0, 1e-8 };
    double eps = 1e-15;
    double ok = 2e5, ng = 0;
    while(ok - ng > eps && (ok - ng) > eps * ok){
        t = (ok + ng) * 0.5;
        vector<pt> cand;

        for(int i=0;i<n;++i){
            cand.emplace_back(x[i], y[i]);
            for(int j=0;j<n;++j){
                auto p = getpt(i, j);
                if(p.first){
                    cand.push_back(p.second.first);
                    cand.push_back(p.second.first + p.second.second);
                    cand.push_back(p.second.first - p.second.second);
                }
            }
        }
        for(auto point: cand){
            if(check(point)) goto l_ok;
            if(check(point+dx)) goto l_ok;
            if(check(point+dy)) goto l_ok;
            if(check(point-dx)) goto l_ok;
            if(check(point-dy)) goto l_ok;
        }
        ng = t;
        continue;
        l_ok:
        ok = t;
    }
    cout << ok << endl;

    return 0;
}