#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, k;
double x[65], y[65], c[65], d[65], dx, dy, R, nx, ny, mx, my, p, l = 0, r = 200000, m, eps = 1e-8;
int main(){
	scanf("%d%d", &n, &k);
	rep(i,n) scanf("%lf%lf%lf", x+i, y+i, c+i);
	if(k == 1){
		puts("0");
		return 0;
	}
	while(eps < r - l){
		bool ok = false;
		m = (l + r) / 2;
		rep(i,n) d[i] = m / c[i];
		rep(i,n) rep(j,i){
			dx = x[j] - x[i];
			dy = y[j] - y[i];
			R = hypot(dx,dy);
			if(R >= d[i] + d[j]) continue;
			if(d[i] > d[j] + R){
				nx = R;
				ny = 0;
			}else if(d[j] > d[i] + R){
				nx = 0;
				ny = 0;
			}else{
				nx = (pow(d[i],2) - pow(d[j],2) + pow(R,2)) / (2 * R);
				ny = sqrt(pow(d[i],2) - pow(nx,2));
			}
			dx /= R; dy /= R;
			rep(kk,2){
				mx = dx * nx + dy * ny * (kk ? -1 : 1) + x[i];
				my = dy * nx + dx * ny * (kk ? 1 : -1) + y[i];
				int count = 0;
				rep(l,n) if(hypot(mx-x[l],my-y[l]) <= d[l]+eps) count++;
				if(count >= k) ok = true;
			}
		}
		(ok ? r : l) = m;
	}
	printf("%.8f\n", l);
}
### 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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, k;
double x[65], y[65], c[65], d[65], dx, dy, R, nx, ny, mx, my, p, l = 0, r = 200000, m, eps = 1e-8;
int main(){
	scanf("%d%d", &n, &k);
	rep(i,n) scanf("%lf%lf%lf", x+i, y+i, c+i);
	if(k == 1){
		puts("0");
		return 0;
	}
	while(eps < r - l){
		bool ok = false;
		m = (l + r) / 2;
		rep(i,n) d[i] = m / c[i];
		rep(i,n) rep(j,i){
			dx = x[j] - x[i];
			dy = y[j] - y[i];
			R = hypot(dx,dy);
			if(R >= d[i] + d[j]) continue;
			if(d[i] > d[j] + R){
				nx = R;
				ny = 0;
			}else if(d[j] > d[i] + R){
				nx = 0;
				ny = 0;
			}else{
				nx = (pow(d[i],2) - pow(d[j],2) + pow(R,2)) / (2 * R);
				ny = sqrt(pow(d[i],2) - pow(nx,2));
			}
			dx /= R; dy /= R;
			rep(kk,2){
				mx = dx * nx + dy * ny * (kk ? -1 : 1) + x[i];
				my = dy * nx + dx * ny * (kk ? 1 : -1) + y[i];
				int count = 0;
				rep(l,n) if(hypot(mx-x[l],my-y[l]) <= d[l]+eps) count++;
				if(count >= k) ok = true;
			}
		}
		(ok ? r : l) = m;
	}
	printf("%.8f\n", l);
}
```