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 <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>
#include <random>

using namespace std;

typedef long double ld;

const ld pi = acos(-1);
const ld eps = 1e-9;

struct Point {
	ld x, y;

	Point() {}

	Point (ld x_, ld y_) {
		x = x_;
		y = y_;
	}

	Point rotate() const {
		return Point(-y, x);
	}

	Point normalize(ld len) const {
		ld c = sqrt(x * x + y * y);
		return Point(x * len / c, y * len / c);
	}

	Point operator+(const Point &other) const {
		return Point(x + other.x, y + other.y);
	}

	Point operator-(const Point &other) const {
		return Point(x - other.x, y - other.y);
	}

	ld dist (const Point &other) const {
		return hypot(x - other.x,  y - other.y);
	}
};

int n, k;
vector<Point> p;
vector<ld> c;
vector<ld> r;

int ans = 0;

void relax(Point to) {
	int u = 0;
	for (int i = 0; i < n; i++) {
		if (p[i].dist(to) < r[i] + eps) u++;
	}
	ans = max(ans, u);
} 

ld msqrt(ld x) {
	if (x < 0) return 0;
	else return sqrt(x);
}

pair<Point, Point> inter(const Point &A, const Point &B, ld ra, ld rb) {
	ld d = A.dist(B);
	ld a = (ra * ra - rb * rb + d * d) / 2 / d;
	ld h = sqrt(ra * ra - a * a);
	Point v = B - A;
	Point u = v.rotate();
	v = v.normalize(a);
	u = u.normalize(h);
	Point H = A + v;
	return {H - u, H + u};
}

void solve() {
	ans = 0;
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			if (p[i].dist(p[j]) > r[i] + r[j] + eps) continue;
			if (p[i].dist(p[j]) + eps < fabs(r[i] - r[j])) continue;
			auto t = inter(p[i], p[j], r[i], r[j]);
			relax(t.first);
			relax(t.second);
		}
		relax(p[i]);
	}
}

int check(ld t) {
	for (int i = 0; i < n; i++) {
		r[i] = t / c[i];
	}
	solve();
	return ans >= k;
}


signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	cout << fixed << setprecision(16);

	cin >> n >> k;
	p.resize(n);
	c.resize(n);
	r.resize(n);
	for (int i = 0; i < n; i++) {
		cin >> p[i].x >> p[i].y >> c[i];
	}
	ld ll = 0;
	ld rr = 1e18 + 239;
	for (int i = 0; i < 100; i++) {
		ld m = (rr + ll) / 2;
		if (check(m)) rr = m;
		else ll = m;
	}
	cout << rr << endl;
}