#include <bits/stdc++.h>
#define X first
#define Y second
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define ini(x, y) memset(x, y, sizeof(x))
#define endl '\n'
#define fastio cin.sync_with_stdio(false); cin.tie(nullptr)
using namespace std;

using ll = long long;
using pii = pair<int, int>;
using pdd = pair<double, double>;
const int MOD = 1e9 + 7;
const int dx[] = { -1, 0, 1, 0, -1, 1, 1, -1 };
const int dy[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

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

double dist(const pdd &a, const pdd &b) {
	double xx = a.X - b.X;
	double yy = a.Y - b.Y;
	return sqrt(xx * xx + yy * yy);
}
vector<pair<double, double>> inter(int a, int b) {
	vector<pair<double, double>> ret;
	double d = dist(pdd(x[a], y[a]), pdd(x[b], y[b]));

	if (r[a] + r[b] < d) return ret;
	if (x[a] == x[b] && y[a] == y[b] || d < abs(r[a] - r[b])) return { { x[a], y[a] } };

	double tmp = (r[a] * r[a] - r[b] * r[b] + d * d) / (2 * d);
	double h = sqrt(r[a] * r[a] - tmp * tmp);

	double xx = x[a] + tmp / d * (x[b] - x[a]);
	double yy = y[a] + tmp / d * (y[b] - y[a]);

	ret.emplace_back(xx + h * (y[b] - y[a]) / d, yy - h * (x[b] - x[a]) / d);
	ret.emplace_back(xx - h * (y[b] - y[a]) / d, yy + h * (x[b] - x[a]) / d);

	return ret;
}
int check(double t) {
	for (int i = 0; i < N; ++i) r[i] = t / c[i];

	int ret = 0;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			auto pts = inter(i, j);
			for (auto &p : pts) {
				int cnt = 0;
				for (int k = 0; k < N; ++k)
					if (dist(p, pdd(x[k], y[k])) <= r[k] + 1e-9) ++cnt;
				ret = max(ret, cnt);
			}
		}
	}

	return ret;
}
int main() {
	fastio;
	cin >> N >> K;
	for (int i = 0; i < N; ++i) cin >> x[i] >> y[i] >> c[i];

	double s = 0, e = 1e10;
	for (int i = 0; i < 100; ++i) {
		double m = (s + e) / 2.0;
		
		if (check(m) < K) s = m;
		else e = m;
	}

	cout << fixed << setprecision(9) << (s + e) / 2.0;

	return 0;
}
### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define X first
#define Y second
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define ini(x, y) memset(x, y, sizeof(x))
#define endl '\n'
#define fastio cin.sync_with_stdio(false); cin.tie(nullptr)
using namespace std;

using ll = long long;
using pii = pair<int, int>;
using pdd = pair<double, double>;
const int MOD = 1e9 + 7;
const int dx[] = { -1, 0, 1, 0, -1, 1, 1, -1 };
const int dy[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

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

double dist(const pdd &a, const pdd &b) {
	double xx = a.X - b.X;
	double yy = a.Y - b.Y;
	return sqrt(xx * xx + yy * yy);
}
vector<pair<double, double>> inter(int a, int b) {
	vector<pair<double, double>> ret;
	double d = dist(pdd(x[a], y[a]), pdd(x[b], y[b]));

	if (r[a] + r[b] < d) return ret;
	if (x[a] == x[b] && y[a] == y[b] || d < abs(r[a] - r[b])) return { { x[a], y[a] } };

	double tmp = (r[a] * r[a] - r[b] * r[b] + d * d) / (2 * d);
	double h = sqrt(r[a] * r[a] - tmp * tmp);

	double xx = x[a] + tmp / d * (x[b] - x[a]);
	double yy = y[a] + tmp / d * (y[b] - y[a]);

	ret.emplace_back(xx + h * (y[b] - y[a]) / d, yy - h * (x[b] - x[a]) / d);
	ret.emplace_back(xx - h * (y[b] - y[a]) / d, yy + h * (x[b] - x[a]) / d);

	return ret;
}
int check(double t) {
	for (int i = 0; i < N; ++i) r[i] = t / c[i];

	int ret = 0;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			auto pts = inter(i, j);
			for (auto &p : pts) {
				int cnt = 0;
				for (int k = 0; k < N; ++k)
					if (dist(p, pdd(x[k], y[k])) <= r[k] + 1e-9) ++cnt;
				ret = max(ret, cnt);
			}
		}
	}

	return ret;
}
int main() {
	fastio;
	cin >> N >> K;
	for (int i = 0; i < N; ++i) cin >> x[i] >> y[i] >> c[i];

	double s = 0, e = 1e10;
	for (int i = 0; i < 100; ++i) {
		double m = (s + e) / 2.0;
		
		if (check(m) < K) s = m;
		else e = m;
	}

	cout << fixed << setprecision(9) << (s + e) / 2.0;

	return 0;
}
```