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:
// 解答不能.
// foryou1006ll氏.
// https://atcoder.jp/contests/abc157/submissions/10562367
#include <bits/stdc++.h>
using namespace std;
#define repex(i, a, b, c) for(int i = a; i < b; i += c)
#define repx(i, a, b) repex(i, a, b, 1)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)
const double eps = 1e-6;
vector<tuple<int, int, int>> info;

double nth_distance(double x, double y, int N, int K){
    vector<double> dist(N);
    rep(i, N) dist[i] = get<2>(info[i]) * hypot(x - get<0>(info[i]), y - get<1>(info[i]));
    nth_element(dist.begin(), dist.begin() + K - 1, dist.end());
    return dist[K - 1];
}

double nth_distance_lb(double x, double y, double r, int N, int K){
    vector<double> dist(N);
    rep(i, N){
        dist[i] = get<2>(info[i]) * hypot(max(abs(x - get<0>(info[i])) - r, 0.0), max(abs(y - get<1>(info[i])) - r, 0.0));
    }
    nth_element(dist.begin(), dist.begin() + K - 1, dist.end());
    return dist[K - 1];
}

int main(){
    int N, K;
    scanf("%d %d", &N, &K);
    rep(i, N){
        int x, y, c;
        scanf("%d %d %d", &x, &y, &c);
        info.emplace_back(x, y, c);
    }
    double ans = 1e9;
    queue<tuple<double, double, double>> q;
    q.emplace(0, 0, 1000);
    while(!q.empty()){
        double x, y, r;
        tie(x, y, r) = q.front();
        q.pop();
        double lb = nth_distance_lb(x, y, r, N, K);
        if(ans - lb > eps * max(1.0, lb)){
            ans = min(ans, nth_distance(x, y, N, K));
            for(double dx : {-0.5, 0.5}){
                for(double dy : {-0.5, 0.5}){
                    q.emplace(x + dx * r, y + dy * r, r / 2.0);
                }
            }
        }
    }
    printf("%.15lf\n", ans);
    return 0;
}