Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 3
-1 0 3
0 0 3
1 0 2
1 1 40

Sample Output 1

2.4

If we put the heat source at \left(-0.2, 0\right), the 1-st, 2-nd, and 3-rd pieces of meat will be ready to eat within 2.4 seconds. This is the optimal place to put the heat source.

Sample Input 2

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

Sample Output 2

7411.2252

Solution in C++14 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#pragma GCC optimize ("Ofast")
#pragma GCC target ("avx")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<unordered_map>
#include<climits>
#include<fstream>
#include<time.h>
#include<cassert>
#include<functional>
#include<numeric>
#include<tuple>

using namespace std;
using ll = long long;
using ld = long double;
#define int long long
#define all(a) (a).begin(),(a).end()
#define EPS (1e-6)
#define fs first
#define sc second
#define xx first
#define yy second.first
#define zz second.second
#define H pair<int, int>
#define P pair<int, pair<int, int>>
#define Q(i,j,k) mkp(i,mkp(j,k))
#define R pair<pair<int, int>, pair<int, int>>
#define S(i,j,k,l) mkp(mkp(i,j),mkp(k,l))
#define rep(i,n) for(int (i) = 0 ; (i) < (n) ; (i)++)
#define rng(i,s,n) for(int (i) = (s) ; (i) < (n) ; (i)++)
#define req(n,i) for(int (i) = (n) - 1 ; (i) >= 0 ; (i)--)
#define range(i,v) for(auto& (i) : v)
#define mkp make_pair
#define mem(x,k) memset(x,k,sizeof(x))
#define vec vector
#define pb emplace_back
#define lb lower_bound
#define ub upper_bound
#define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end())
#define getidx(b,i) lower_bound(all(b),(i))-b.begin()
#define ssp(i,n) (i==n-1?"\n":" ")
#define ctoi(c) (int)(c-'0')
#define itoc(c) (char)(c+'0')
#define itn int
#define pri_que priority_queue
#define pp(x,y) pb(H{x,y})
#define ppp(x,y,z) pb(Q(x,y,z))
#define pppp(x,y,z,w) pb(S(x,y,z,w))
#define cyes cout<<"Yes\n"
#define cno cout<<"No\n"
#define endl "\n"
const int mod = 1e9 + 7;
const int Mod = 998244353;
const ll inf = 3 * 1e18;
const int Inf = 15 * 1e8;
const vec<int>dx{ -1,1,0,0 }, dy{ 0,0,-1,1 };
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
ll read() { ll u, k = scanf("%lld", &u); return u; }
string reads() { string s; cin >> s; return s; }
H readh(bool g = 0) { H u; ll k = scanf("%lld %lld", &u.fs, &u.sc); if (g) u.fs--, u.sc--; return u; }
void printh(H t) { printf("%lld %lld\n", t.fs, t.sc); }
bool inarea(H t, int h, int w) { return 0 <= t.fs && t.fs < h && 0 <= t.sc && t.sc < w; }
ll gcd(ll i, ll j) { return j ? gcd(j, i % j) : i; }
ll mod_pow(ll x, ll n, ll p) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % p;
		x = x * x % p;
		n >>= 1;
	}
	return res;
}//x^n%p
ll bitcount(ll x) {
	int sum = 0; for (int i = 0; i < 60; i++)if ((1ll << i) & x) sum++;
	return sum;
}
/*const int fn_ = 1000005;
ll fact_[fn_], comp_[fn_];
ll comb(ll x, ll y, ll Mod = mod) {
	if (!fact_[0]) {
		fact_[0] = 1; comp_[0] = 1;
		for (int i = 1; i < fn_; i++) {
			fact_[i] = fact_[i - 1] * i % Mod;
			comp_[i] = mod_pow(fact_[i], Mod - 2, Mod);
		}
	}
	if (x < y) return 0;
	return fact_[x] * comp_[x - y] % Mod * comp_[y] % Mod;
}*/
//--------------------------------------------------------------
int n, k;
#define T pair<ld, ld>
vec<P>a;
vec<T>v;
inline ld dist(T a, T b) {
	return sqrt((a.fs - b.fs) * (a.fs - b.fs) + (a.sc - b.sc) * (a.sc - b.sc));
}
vec<T>se;
ld d, rc, rs;
T diff;
int res, sum;
T g;
int max_en(ld t) {
	se.clear();
	T k;
	rep(i, n) {
		rng(j, i + 1, n) {
			ld s = t / (ld)a[i].zz, r = t / (ld)a[j].zz;
			if (dist(v[i], v[j]) < s + r + EPS) {
				k = T{ v[j].fs - v[i].fs,v[j].sc - v[i].sc };
				d = dist(v[i], v[j]);
				rc = (d * d / (2 * d) + s * s / (2 * d) - r * r / (2 * d));
				rs = sqrt(s * s - rc * rc);
				diff.fs = (v[j].fs - v[i].fs) / d;
				diff.sc = (v[j].sc - v[i].sc) / d;
				se.pb(T{ v[i].fs + rc * diff.fs + rs * diff.sc,v[i].sc + rc * diff.sc - rs * diff.fs });
				se.pb(T{ v[i].fs + rc * diff.fs - rs * diff.sc,v[i].sc + rc * diff.sc + rs * diff.fs });
			}
		}
		se.pb(v[i]);
	}
	res = 0;
	rep(_,se.size()){
		g = se[_];
		sum = 0;
		rep(i, n) {
			if (dist(v[i], g) < t / a[i].zz + EPS) {
				sum++;
			}
		}
		res = max(res, sum);
	}
	return res;
}
signed main() {
	cin >> n >> k;
	int rr = 0, x = -inf, y = -inf;
	rep(i, n) {
		int u, c, r;
		scanf("%lld %lld %lld", &u, &c, &r);
		a.ppp(u, c, r);
		rr = max(rr, a[i].zz);
		x = max(x, abs(a[i].xx));
		y = max(y, abs(a[i].yy));
	}
	crdcomp(a);
	n = a.size();
	rep(i, n) v.pb(a[i].xx, a[i].yy);
	ld g = 0;
	g = 1e-7;
	ld ok = rr * (x + y), ng = 0, mid;
	while (ok - ng > g) {
		mid = (ok + ng) / 2;
		if (max_en(mid) >= k) {
			ok = mid;
		}
		else {
			ng = mid;
			g = max(g, ng * 1e-6);
		}
	}
	printf("%.7LF\n", ok);
}