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 <cassert>
#include <iomanip>
#include <functional>
#include <cmath>
#include <iostream>
using namespace std;

int main() {
    int n, k; cin >> n >> k;
    int x[n], y[n], c[n];
    for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> c[i];
    if (k == 1) { cout << "0\n"; return 0; }
    double lb = 0, ub = 2e7;
    function<bool(double, double, double)> check = [&](double cx, double cy, double v) {
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            double a = cx - x[i], b = cy - y[i];
            if (c[i] * sqrt(a * a + b * b) <= v * (1 + 1e-7)) cnt++;
        }
        return cnt >= k;
    };
    for (int xx = 0; xx < 100; xx++) {
        double v = (lb + ub) / 2;
        bool ok = false;
        for (int i = 0; !ok && i < n; i++) {
            if (check(x[i], y[i], v)) {
                ok = true;
                break;
            }
            for (int j = i+1; j < n; j++) {
                double vx = x[j] - x[i], vy = y[j] - y[i],
                    d = sqrt(vx * vx + vy * vy),
                    di = v / c[i], dj = v / c[j];
                if (d > di + dj || di > d + dj || dj > d + di) continue;
                double k = (d * d - dj * dj + di * di) / 2 / d,
                    h = sqrt(di * di - k * k),
                    cx1 = x[i] + k * vx / d - h * vy / d,
                    cx2 = x[i] + k * vx / d + h * vy / d,
                    cy1 = y[i] + k * vy / d + h * vx / d,
                    cy2 = y[i] + k * vy / d - h * vx / d;
                // cerr << d << ' ' << di << ' ' << dj << ' ' << k << ' ' << h << ' ' << cx1 << ' ' << cx2 << ' ' << cy1 << ' ' << cy2 << endl;
                
                // cerr<< (abs(sqrt(pow(cx1 - x[i], 2) + pow(cy1 - y[i], 2)) - di)) << ' '
                //     <<  (abs(sqrt(pow(cx2 - x[i], 2) + pow(cy2 - y[i], 2)) - di)) << ' '
                //     <<  (abs(sqrt(pow(cx2 - x[j], 2) + pow(cy2 - y[j], 2)) - dj)) << endl;

                if (check(cx1, cy1, v) || check(cx2, cy2, v)) {
                    ok = true;
                    break;
                }
            }
        }
        if (ok) ub = v;
        else lb = v;
        // cerr << setprecision(12) << v << ' ' << (ok ? "OK" : "NG") << endl;
    }
    cout << setprecision(12) << lb << endl;
}