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", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T &t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T &t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
const long long MOD2 = (long long)998244353 * (long long)998244353;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 998244353) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
struct Point {
  long double x, y;
  Point() {}
  Point(long double x, long double y) : x(x), y(y) {}
  long double abs() const { return hypot(x, y); }
  long double arg() const { return atan2(y, x); }
  Point operator*(double o) const { return Point(x * o, y * o); }
  Point operator+(const Point &o) const { return Point(x + o.x, y + o.y); }
  Point operator-(const Point &o) const { return Point(x - o.x, y - o.y); }
  Point operator/(double o) const { return Point(x / o, y / o); }
  bool operator<(const Point &o) const {
    return x < o.x - 1e-9 || (x < o.x + 1e-9 && y < o.y - 1e-9);
  }
  friend bool operator==(const Point &r1, const Point &r2) {
    return r1.x == r2.x && r1.y == r2.y;
  }
  Point scale(double o) const { return *this * (o / abs()); }
  Point rotY() const { return Point(-y, x); }
  Point rotX() const { return Point(y, -x); }
  long double cross(Point b) const { return x * b.y - b.x * y; }
  long double dot(Point b) const { return x * b.x + y * b.y; }
  long double disToSeg(Point &a, Point &b) {
    return fabs(((*this) - a).cross(b - a) / (b - a).abs());
  }
  pair<long long, long long> getVec() {
    if (x == 0) return {0, y > 0 ? 1 : -1};
    if (y == 0) return {x < 0 ? -1 : 1, 0};
    long long gg = gcd1(::abs(x), ::abs(y));
    long long px = x / gg, py = y / gg;
    return {px, py};
  }
  void readin() { scanf("%Lf%Lf", &x, &y); }
} p[(100035)];
long double dis[(100035)], the[(100035)];
int b[(100035) * 2], id[(100035)];
void update(int x, int v, int sz) {
  for (int i = x; i <= sz; i += i & -i) {
    b[i] += v;
  }
}
int query(int x) {
  int ret = 0;
  for (; x > 0; ret += b[x], x -= x & -x)
    ;
  return ret;
}
bool ck(long double mid) {
  int inc = 0;
  vector<pair<long double, long double>> v;
  vector<pair<long double, int>> vp;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    if (dis[i] <= mid)
      inc++;
    else {
      long double det = acos(mid / dis[i]);
      long double L = the[i] - det, R = the[i] + det;
      if (L < -acos(-1)) L += acos(-1) + acos(-1);
      if (R > acos(-1)) R -= acos(-1) + acos(-1);
      if (L > R) swap(L, R);
      v.push_back({L, R});
    }
  }
  long long ans = (long long)inc * (inc - 1) / 2 + (long long)inc * (n - inc);
  if (ans >= lk) return 1;
  memset(b, 0, sizeof b);
  int SZ = v.size();
  map<long double, int> mp;
  int cnt = 0;
  for (int(i) = 0; (i) < (int)(SZ); (i)++) {
    mp[v[i].first];
    vp.push_back({v[i].first, -(i + 1)});
    vp.push_back({v[i].second, i + 1});
  }
  sort(vp.begin(), vp.end());
  for (auto &p : mp) p.second = ++cnt;
  for (int(i) = 0; (i) < (int)(SZ); (i)++) {
    id[i] = mp[v[i].first];
  }
  int N = vp.size();
  long long bd = 0;
  for (int(i) = 0; (i) < (int)(N); (i)++) {
    int w = abs(vp[i].second) - 1;
    if (vp[i].second < 0) {
      update(id[w], 1, cnt);
    } else {
      update(id[w], -1, cnt);
      bd += query(cnt) - query(id[w]);
    }
  }
  ans += (long long)(n - inc) * (n - inc - 1) / 2 - bd;
  return ans >= lk;
}
void fmain(int tid) {
  scanf("%d%lld", &n, &lk);
  for (int(i) = 1; (i) <= (int)(n); (i)++) p[i].readin();
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    dis[i] = p[i].abs();
    the[i] = p[i].arg();
  }
  long double l = 0, r = 2e8;
  while (abs(l - r) > 1e-9) {
    long double mid = (l + r) / 2;
    if (ck(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%.10Lf\n", l);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}