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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) begin(v),end(v)
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
using ll = long long;
using pii = pair<int, int>;
constexpr ll INF = 1ll<<30;
constexpr ll longINF = 1ll<<60;
constexpr ll MOD = 1000000007;
constexpr bool debug = 0;
//---------------------------------//

constexpr double EPS = 1e-6;

int N, K;
int x[60], y[60], c[60];

bool solve2(double X, double Y, double M) {
	int cnt = 0;

	REP(i, N) {
		double cur = c[i]*c[i] * (pow(x[i]-X,2)+pow(y[i]-Y,2));
		if (sqrt(cur) <= M + EPS) ++cnt;
	}
	return cnt >= K;
}

bool solve(double X) {
	// x[i], y[i]中心
	vector<double> xp, yp;
	REP(i, N) FOR(j, i + 1, N) { // iとjの中心 iからのj相対を
		double x1 = x[j] - x[i], y1 = y[j] - y[i];
		
		double r1 = X / c[i];
		double r2 = X / c[j];
		
		if (x1*x1 + y1*y1 > pow(r1+r2, 2) - EPS) continue;
		
		double a = (x1*x1 + y1*y1 + r1*r1 - r2*r2) / 2;
		
		xp.push_back((a*x1 + y1*sqrt((x1*x1+y1*y1)*r1*r1-a*a)) / (x1*x1+y1*y1) + x[i]);
		yp.push_back((a*y1 - x1*sqrt((x1*x1+y1*y1)*r1*r1-a*a)) / (x1*x1+y1*y1) + y[i]);
		xp.push_back((a*x1 - y1*sqrt((x1*x1+y1*y1)*r1*r1-a*a)) / (x1*x1+y1*y1) + x[i]);
		yp.push_back((a*y1 + x1*sqrt((x1*x1+y1*y1)*r1*r1-a*a)) / (x1*x1+y1*y1) + y[i]);
	}
	
	REP(i, N) {
		xp.push_back(x[i]);
		yp.push_back(y[i]);
	}
	
	bool ok = false;
	REP(i, xp.size()) if (solve2(xp[i], yp[i], X)) ok = true;
	return ok;
}

int main() {
	cin >> N >> K;
	REP(i, N) scanf("%d %d %d", x + i, y + i, c + i);
	
	double l = 0, r = INF;
	REP(t, 60) {
		double m = (l + r) / 2;
		if (solve(m)) r = m;
		else l = m;
	}
	
	printf("%.17lf\n", l);
	
	return 0;
}