#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
typedef long long ll;
typedef long double ld;
using namespace std;

ld x[70], y[70], c[70];
ld r[70], d[70][70], ang[70][70];
ld pi=abs(acos(-1.0));

ld dist(ld x1, ld y1, ld x2, ld y2){
	return pow((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2), 0.5);
}

ld yogen(ld a, ld b, ld c){
	return acos((a*a+b*b-c*c)/(a*b*2.0));
}

int main() {
	int N, K;
	cin >> N >> K;
	for(int i=0; i<N; ++i){
		cin >> x[i] >> y[i] >> c[i];
	}
	for(int i=0; i<N; ++i){
		for(int j=0; j<N; ++j){
			d[i][j]=dist(x[i], y[i], x[j], y[j]);
			ang[i][j]=atan2(y[j]-y[i], x[j]-x[i]);
		}
	}
	ld ng=0.0, ok=1e9, er=1e-10;
	while(ng+er<ok){
		ld mid=(ng+ok)/2.0;
		for(int i=0; i<N; ++i) r[i]=mid/c[i];
		int mx=0;
		for(int i=0; i<N; ++i){
			vector<pair<ld, int>> v;
			int stt=1;
			for(int j=0; j<N; ++j){
				if(i!=j){
					if(d[i][j]+r[i]<=r[j]+er) ++stt;
					else if(d[i][j]+er<r[i]+r[j]){
						ld p=yogen(r[i], d[i][j], r[j]);
						if(ang[i][j]-p<(-pi)){
							++stt;
							v.emplace_back(make_pair(ang[i][j]-p+pi*2.0, 1));
						}else{
							v.emplace_back(make_pair(ang[i][j]-p, 1));
						}
						if(ang[i][j]+p>pi){
							++stt;
							v.emplace_back(make_pair(ang[i][j]+p-pi*2.0, -1));
						}else{
							v.emplace_back(make_pair(ang[i][j]+p, -1));
						}
					}
				}
			}
			sort(v.begin(), v.end());
			mx=max(mx, stt);
			for(int i=0; i<v.size(); ++i){
				stt += v[i].second;
				mx=max(mx, stt);
			}
		}
		if(mx>=K) ok=mid;
		else ng=mid;
	}
	cout << setprecision(15) << ok << endl;
	return 0;
}
### Prompt

Create a solution in cpp for 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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
typedef long long ll;
typedef long double ld;
using namespace std;

ld x[70], y[70], c[70];
ld r[70], d[70][70], ang[70][70];
ld pi=abs(acos(-1.0));

ld dist(ld x1, ld y1, ld x2, ld y2){
	return pow((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2), 0.5);
}

ld yogen(ld a, ld b, ld c){
	return acos((a*a+b*b-c*c)/(a*b*2.0));
}

int main() {
	int N, K;
	cin >> N >> K;
	for(int i=0; i<N; ++i){
		cin >> x[i] >> y[i] >> c[i];
	}
	for(int i=0; i<N; ++i){
		for(int j=0; j<N; ++j){
			d[i][j]=dist(x[i], y[i], x[j], y[j]);
			ang[i][j]=atan2(y[j]-y[i], x[j]-x[i]);
		}
	}
	ld ng=0.0, ok=1e9, er=1e-10;
	while(ng+er<ok){
		ld mid=(ng+ok)/2.0;
		for(int i=0; i<N; ++i) r[i]=mid/c[i];
		int mx=0;
		for(int i=0; i<N; ++i){
			vector<pair<ld, int>> v;
			int stt=1;
			for(int j=0; j<N; ++j){
				if(i!=j){
					if(d[i][j]+r[i]<=r[j]+er) ++stt;
					else if(d[i][j]+er<r[i]+r[j]){
						ld p=yogen(r[i], d[i][j], r[j]);
						if(ang[i][j]-p<(-pi)){
							++stt;
							v.emplace_back(make_pair(ang[i][j]-p+pi*2.0, 1));
						}else{
							v.emplace_back(make_pair(ang[i][j]-p, 1));
						}
						if(ang[i][j]+p>pi){
							++stt;
							v.emplace_back(make_pair(ang[i][j]+p-pi*2.0, -1));
						}else{
							v.emplace_back(make_pair(ang[i][j]+p, -1));
						}
					}
				}
			}
			sort(v.begin(), v.end());
			mx=max(mx, stt);
			for(int i=0; i<v.size(); ++i){
				stt += v[i].second;
				mx=max(mx, stt);
			}
		}
		if(mx>=K) ok=mid;
		else ng=mid;
	}
	cout << setprecision(15) << ok << endl;
	return 0;
}
```