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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <functional>
#include <bitset>
#include <numeric>
#include <cassert>
#include <random>
using namespace std;
using ll = long long;


int main() {
	int n, k; cin >> n >> k;
	vector<double> x(n), y(n), c(n);
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i] >> c[i];
	}
	auto dist = [](double a, double b, double c = 0, double d = 0) {
		return sqrt((a - c)*(a - c) + (b - d)*(b - d));
	};
	auto cntp = [&](double u, double v, double r) {
		int cnt = 0;
		for (int i = 0; i < n; i++) {
			double d = dist(x[i], y[i], u, v);
			cnt += c[i]*d <= r + 1e-8;
		}
		return cnt;
	};
	auto cross = [&](double a, double b, double r, double c, double d, double s, bool neg) -> pair<double, double> {
		c -= a;	d -= b;
		double dcd = dist(c, d)*dist(c, d);
		double t = 0.5 * (dcd + r*r - s*s);
		double sgn = neg ? -1 : 1;
		if (dcd*r*r - t*t < 0) {
			return make_pair(1e18, 1e18);
		}
		double xx = (t*c + sgn*d*sqrt(dcd*r*r - t*t)) / dcd;
		double yy = (t*d - sgn*c*sqrt(dcd*r*r - t*t)) / dcd;
		xx += a; yy += b;
		return make_pair(xx, yy);
	};
	double lo = 0, hi = 1e9;
	for (int iter = 0; iter < 100; iter++) {
		double mid = (lo + hi) / 2;
		bool flag = false;
		for (int i = 0; i < n; i++) {
			flag |= cntp(x[i], y[i], mid) >= k;
		}
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				double u, v;
				tie(u, v) = cross(x[i], y[i], mid/c[i], x[j], y[j], mid/c[j], false);
				if (u < 1e17) flag |= cntp(u, v, mid) >= k;
				tie(u, v) = cross(x[i], y[i], mid/c[i], x[j], y[j], mid/c[j], true);
				if (u < 1e17) flag |= cntp(u, v, mid) >= k;
			}
		}
		if (flag) hi = mid;
		else lo = mid;
	}
	printf("%.10f\n", lo);
	return 0;
}