Problem: You are given an integer k and n distinct points with integer coordinates on the Euclidean plane, the i-th point has coordinates (x_i, y_i).

Consider a list of all the (n(n - 1))/(2) pairs of points ((x_i, y_i), (x_j, y_j)) (1 ≤ i < j ≤ n). For every such pair, write out the distance from the line through these two points to the origin (0, 0).

Your goal is to calculate the k-th smallest number among these distances.

Input

The first line contains two integers n, k (2 ≤ n ≤ 10^5, 1 ≤ k ≤ (n(n - 1))/(2)).

The i-th of the next n lines contains two integers x_i and y_i (-10^4 ≤ x_i, y_i ≤ 10^4) — the coordinates of the i-th point. It is guaranteed that all given points are pairwise distinct.

Output

You should output one number — the k-th smallest distance from the origin. Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Example

Input


4 3
2 1
-2 -1
0 -1
-2 4


Output


0.707106780737

Note

There are 6 pairs of points: 

  * Line 1-2 : distance 0 from the origin 
  * Line 1-3 : distance \frac{√{2}}{2} ≈ 0.707106781 from the origin 
  * Line 1-4 : distance 2 from the origin 
  * Line 2-3 : distance 1 from the origin 
  * Line 2-4 : distance 2 from the origin 
  * Line 3-4 : distance \frac{2}{√{29}} ≈ 0.371390676 from the origin 

The third smallest distance among those is approximately 0.707106781.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
void ckmn(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void ckmx(T& a, T b) {
  a = max(a, b);
}
void rd(int& x) { scanf("%i", &x); }
void rd(long long& x) { scanf("%lld", &x); }
void rd(char* x) { scanf("%s", x); }
void rd(double& x) { scanf("%lf", &x); }
void rd(string& x) { scanf("%s", &x); }
template <typename T1, typename T2>
void rd(pair<T1, T2>& x) {
  rd(x.first);
  rd(x.second);
}
template <typename T>
void rd(vector<T>& x) {
  for (T& i : x) rd(i);
}
template <typename T, typename... A>
void rd(T& x, A&... args) {
  rd(x);
  rd(args...);
}
template <typename T>
void rd() {
  T x;
  rd(x);
  return x;
}
int ri() {
  int x;
  rd(x);
  return x;
}
template <typename T>
vector<T> rv(int n) {
  vector<T> x(n);
  rd(x);
  return x;
}
template <typename T>
void ra(T a[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]);
}
template <typename T1, typename T2>
void ra(T1 a[], T2 b[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]);
}
template <typename T1, typename T2, typename T3>
void ra(T1 a[], T2 b[], T3 c[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]), rd(c[st + i]);
}
void re(vector<int> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    rd(u, v);
    E[u].push_back(v);
    if (!dir) E[v].push_back(u);
  }
}
template <typename T>
void re(vector<pair<int, T>> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    T w;
    rd(u, v, w);
    E[u].push_back({v, w});
    if (!dir) E[v].push_back({u, w});
  }
}
const int N = 100050;
const double PI = acos(-1);
struct pt {
  double x, y;
  pt() {}
  pt(double a, double b) : x(a), y(b) {}
} pts[N];
double abs(pt p) { return sqrt(p.x * p.x + p.y * p.y); }
void fix(double& x) {
  while (x > PI) x -= 2 * PI;
  while (x <= -PI) x += 2 * PI;
}
pair<double, double> Make(pt p, double r) {
  double dist = abs(p);
  double phi = acos(r / dist);
  double alpha = atan2(p.y, p.x);
  double L = alpha + phi;
  fix(L);
  double R = alpha - phi;
  fix(R);
  if (L > R) swap(L, R);
  return {L, R};
}
const int M = 80 * N;
int root[N * 2], ls[M], rs[M], tsz, sum[M];
void Set(int p, int& c, int ss, int se, int qi) {
  c = ++tsz;
  ls[c] = ls[p];
  rs[c] = rs[p];
  sum[c] = sum[p] + 1;
  if (ss == se) return;
  int mid = ss + se >> 1;
  if (qi <= mid)
    Set(ls[p], ls[c], ss, mid, qi);
  else
    Set(rs[p], rs[c], mid + 1, se, qi);
}
int Get(int p, int c, int ss, int se, int qs, int qe) {
  if (qs > qe || qs > se || ss > qe) return 0;
  if (qs <= ss && qe >= se) return sum[c] - sum[p];
  int mid = ss + se >> 1;
  return Get(ls[p], ls[c], ss, mid, qs, qe) +
         Get(rs[p], rs[c], mid + 1, se, qs, qe);
}
void cl() {
  for (int i = 1; i <= tsz; i++) ls[i] = rs[i] = sum[i] = 0;
  tsz = 0;
}
int n;
long long k;
pair<double, double> ang[N];
bool Check(double r) {
  vector<pair<double, int>> all;
  long long cnt = 0;
  int lef = n;
  for (int i = 1; i <= n; i++) {
    double dist = abs(pts[i]);
    if (dist <= r) {
      cnt += lef - 1;
      lef--;
    } else {
      ang[i] = Make(pts[i], r);
      all.push_back({ang[i].first, -i});
      all.push_back({ang[i].second, i});
    }
  }
  sort(all.begin(), all.end());
  cl();
  map<int, int> idx;
  for (int i = 0; i < all.size(); i++) {
    int j = all[i].second;
    root[i + 1] = root[i];
    if (j > 0) {
      Set(root[i + 1], root[i + 1], 1, all.size(), idx[j]);
      cnt += Get(root[idx[j]], root[i + 1], 1, all.size(), idx[j] + 1, i);
      cnt += Get(root[0], root[idx[j]], 1, all.size(), 1, idx[j] - 1);
    } else {
      idx[-j] = i + 1;
    }
  }
  return cnt >= k;
}
int main() {
  rd(n, k);
  for (int i = 1; i <= n; i++) {
    int x, y;
    rd(x, y);
    pts[i] = pt(x, y);
  }
  double top = 1e5, bot = 0;
  while (top - bot > 1e-7) {
    double mid = (top + bot) / 2;
    if (Check(mid))
      top = mid;
    else
      bot = mid;
  }
  printf("%.12f\n", (top + bot) / 2);
  return 0;
}