#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <vector>
#include <math.h>
using namespace std;
const int N = 65;
const double EPS = 1e-9;
int n, k;
double xx[N], yy[N], c[N], r[N];

typedef pair<double, double> PDD;
typedef vector<PDD> VPDD;
typedef double DB;
#define x first
#define y second

DB distance(PDD a, PDD b) {
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

VPDD cal(DB x1, DB y1, DB r1, DB x2, DB y2, DB r2) {
	x1 -= x2, y1 -= y2;
	DB S = x1 * x1 + y1 * y1;
	DB a = (S + r2 * r2 - r1 * r1) / 2;
	DB D = S * r2*r2 - a * a;
	if (D < 0) return {};
	DB A1 = a * x1, B1 = y1 * sqrt(D);
	DB A2 = a * y1, B2 = x1 * sqrt(D);
	return { {(A1 + B1) / S + x2,(A2 - B2) / S + y2},{(A1 - B1) / S + x2,(A2 + B2) / S + y2} };
}

bool check(DB t) {
	for (int i = 0; i < n; i++)
		r[i] = t / c[i];
	VPDD vec;
	for (int i = 0; i < n; i++) {
		vec.push_back({ xx[i],yy[i] });
		for (int j = i + 1; j < n; j++) {
			VPDD v = cal(xx[i], yy[i], r[i], xx[j], yy[j], r[j]);
			for (auto it : v)
				vec.push_back(it);
		}
	}
	for (auto it : vec) {
		int cnt = 0;
		for (int i = 0; i < n; i++)
			if (r[i] + EPS >= distance(it, { xx[i],yy[i] }))
				cnt++;
		if (cnt >= k)
			return true;
	}
	return false;
}

int main() {
	cin >> n >> k;
	for (int i = 0; i < n; i++)
		cin >> xx[i] >> yy[i] >> c[i];
	
	DB left = 0, right = 1e9;
	for (int i = 0; i < 100; i++) {
		DB mid = (left + right) / 2;
		if (check(mid)) right = mid;
		else left = mid;
	}

	printf("%.12f\n", right);

	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <vector>
#include <math.h>
using namespace std;
const int N = 65;
const double EPS = 1e-9;
int n, k;
double xx[N], yy[N], c[N], r[N];

typedef pair<double, double> PDD;
typedef vector<PDD> VPDD;
typedef double DB;
#define x first
#define y second

DB distance(PDD a, PDD b) {
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

VPDD cal(DB x1, DB y1, DB r1, DB x2, DB y2, DB r2) {
	x1 -= x2, y1 -= y2;
	DB S = x1 * x1 + y1 * y1;
	DB a = (S + r2 * r2 - r1 * r1) / 2;
	DB D = S * r2*r2 - a * a;
	if (D < 0) return {};
	DB A1 = a * x1, B1 = y1 * sqrt(D);
	DB A2 = a * y1, B2 = x1 * sqrt(D);
	return { {(A1 + B1) / S + x2,(A2 - B2) / S + y2},{(A1 - B1) / S + x2,(A2 + B2) / S + y2} };
}

bool check(DB t) {
	for (int i = 0; i < n; i++)
		r[i] = t / c[i];
	VPDD vec;
	for (int i = 0; i < n; i++) {
		vec.push_back({ xx[i],yy[i] });
		for (int j = i + 1; j < n; j++) {
			VPDD v = cal(xx[i], yy[i], r[i], xx[j], yy[j], r[j]);
			for (auto it : v)
				vec.push_back(it);
		}
	}
	for (auto it : vec) {
		int cnt = 0;
		for (int i = 0; i < n; i++)
			if (r[i] + EPS >= distance(it, { xx[i],yy[i] }))
				cnt++;
		if (cnt >= k)
			return true;
	}
	return false;
}

int main() {
	cin >> n >> k;
	for (int i = 0; i < n; i++)
		cin >> xx[i] >> yy[i] >> c[i];
	
	DB left = 0, right = 1e9;
	for (int i = 0; i < 100; i++) {
		DB mid = (left + right) / 2;
		if (check(mid)) right = mid;
		else left = mid;
	}

	printf("%.12f\n", right);

	return 0;
}
```