#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-12L;
const long double pi = acosl(-1);
struct point {
  long double x;
  long double y;
  point(long double _x = 0, long double _y = 0) : x(_x), y(_y) {}
  point &operator+=(point other) {
    x += other.x;
    y += other.y;
    return *this;
  }
  point &operator-=(point other) {
    x -= other.x;
    y -= other.y;
    return *this;
  }
  point &operator*=(long double other) {
    x *= other;
    y *= other;
    return *this;
  }
  point &operator/=(long double other) {
    x /= other;
    y /= other;
    return *this;
  }
  point operator-() { return point(*this) *= -1; }
  point operator+(point rhs) { return point(*this) += rhs; }
  point operator-(point rhs) { return point(*this) -= rhs; }
  point operator*(long double rhs) { return point(*this) *= rhs; }
  point operator/(long double rhs) { return point(*this) /= rhs; }
  bool operator==(point rhs) {
    return (abs(x - rhs.x) < eps && abs(y - rhs.y) < eps);
  }
  bool operator!=(point rhs) {
    return (abs(x - rhs.x) > eps || abs(y - rhs.y) > eps);
  }
};
string to_string(point a) {
  return "(" + to_string(a.x) + ", " + to_string(a.y) + ")";
}
long double dot(point a, point b) { return a.x * b.x + a.y * b.y; }
long double cross(point a, point b) { return a.x * b.y - a.y * b.x; }
long double dist2(point a, point b) { return dot(a - b, a - b); }
long double dist(point a, point b) { return sqrtl(dist2(a, b)); }
long double norm(point a) { return sqrtl(dot(a, a)); }
long double arg(point a) { return atan2l(a.y, a.x); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<point> p;
  for (int i = 0; i < n; i++) {
    point q;
    cin >> q.x >> q.y;
    if (abs(q.x) < eps && abs(q.y) < eps) {
      k--;
    } else {
      p.emplace_back(q);
    }
  }
  n = (int)p.size();
  if (k <= 0) {
    cout << 0 << '\n';
    return 0;
  }
  long double lo = 0, hi = 2e5;
  while (hi - lo > 1e-6) {
    long double mid = (hi + lo) / 2;
    vector<long double> add, remove, all;
    int now = 0;
    for (int i = 0; i < n; i++) {
      long double d = norm(p[i]) / 2;
      if (d > mid) {
        continue;
      }
      long double a = sqrtl(mid * mid - d * d);
      long double t = atanl(a / d);
      long double f = atan2l(p[i].y, p[i].x);
      long double beg = f - t;
      long double end = f + t;
      if (beg < -pi) {
        beg += 2 * pi;
      }
      if (end > pi) {
        end -= 2 * pi;
      }
      if (end < beg) {
        now++;
      }
      add.emplace_back(beg);
      remove.emplace_back(end);
      all.emplace_back(beg);
      all.emplace_back(end);
    };
    sort(add.begin(), add.end());
    sort(remove.begin(), remove.end());
    sort(all.begin(), all.end());
    all.resize(unique(all.begin(), all.end()) - all.begin());
    int ok = 0;
    int it1 = 0, it2 = 0;
    if (now >= k) {
      ok = 1;
    }
    for (auto v : all) {
      while (it1 < (int)add.size() && abs(add[it1] - v) < eps) {
        now++;
        it1++;
      }
      while (it2 < (int)remove.size() && abs(remove[it2] - v) < eps) {
        now--;
        it2++;
      }
      if (now >= k) {
        ok = 1;
      }
    }
    if (ok) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  cout << fixed << setprecision(12);
  cout << hi << '\n';
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

Mr. Chanek lives in a city represented as a plane. He wants to build an amusement park in the shape of a circle of radius r. The circle must touch the origin (point (0, 0)).

There are n bird habitats that can be a photo spot for the tourists in the park. The i-th bird habitat is at point p_i = (x_i, y_i). 

Find the minimum radius r of a park with at least k bird habitats inside. 

A point is considered to be inside the park if and only if the distance between p_i and the center of the park is less than or equal to the radius of the park. Note that the center and the radius of the park do not need to be integers.

In this problem, it is guaranteed that the given input always has a solution with r ≤ 2 ⋅ 10^5.

Input

The first line contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ n) — the number of bird habitats in the city and the number of bird habitats required to be inside the park.

The i-th of the next n lines contains two integers x_i and y_i (0 ≤ |x_i|, |y_i| ≤ 10^5) — the position of the i-th bird habitat.

Output

Output a single real number r denoting the minimum radius of a park with at least k bird habitats inside. It is guaranteed that the given input always has a solution with r ≤ 2 ⋅ 10^5.

Your answer is considered correct if its absolute or relative error does not exceed 10^{-4}.

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^{-4}.

Examples

Input


8 4
-3 1
-4 4
1 5
2 2
2 -2
-2 -4
-1 -1
-6 0


Output


3.1622776589


Input


1 1
0 0


Output


0.0000000000

Note

In the first example, Mr. Chanek can put the center of the park at (-3, -1) with radius √{10} ≈ 3.162. It can be proven this is the minimum r.

The following illustrates the first example. The blue points represent bird habitats and the red circle represents the amusement park.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-12L;
const long double pi = acosl(-1);
struct point {
  long double x;
  long double y;
  point(long double _x = 0, long double _y = 0) : x(_x), y(_y) {}
  point &operator+=(point other) {
    x += other.x;
    y += other.y;
    return *this;
  }
  point &operator-=(point other) {
    x -= other.x;
    y -= other.y;
    return *this;
  }
  point &operator*=(long double other) {
    x *= other;
    y *= other;
    return *this;
  }
  point &operator/=(long double other) {
    x /= other;
    y /= other;
    return *this;
  }
  point operator-() { return point(*this) *= -1; }
  point operator+(point rhs) { return point(*this) += rhs; }
  point operator-(point rhs) { return point(*this) -= rhs; }
  point operator*(long double rhs) { return point(*this) *= rhs; }
  point operator/(long double rhs) { return point(*this) /= rhs; }
  bool operator==(point rhs) {
    return (abs(x - rhs.x) < eps && abs(y - rhs.y) < eps);
  }
  bool operator!=(point rhs) {
    return (abs(x - rhs.x) > eps || abs(y - rhs.y) > eps);
  }
};
string to_string(point a) {
  return "(" + to_string(a.x) + ", " + to_string(a.y) + ")";
}
long double dot(point a, point b) { return a.x * b.x + a.y * b.y; }
long double cross(point a, point b) { return a.x * b.y - a.y * b.x; }
long double dist2(point a, point b) { return dot(a - b, a - b); }
long double dist(point a, point b) { return sqrtl(dist2(a, b)); }
long double norm(point a) { return sqrtl(dot(a, a)); }
long double arg(point a) { return atan2l(a.y, a.x); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<point> p;
  for (int i = 0; i < n; i++) {
    point q;
    cin >> q.x >> q.y;
    if (abs(q.x) < eps && abs(q.y) < eps) {
      k--;
    } else {
      p.emplace_back(q);
    }
  }
  n = (int)p.size();
  if (k <= 0) {
    cout << 0 << '\n';
    return 0;
  }
  long double lo = 0, hi = 2e5;
  while (hi - lo > 1e-6) {
    long double mid = (hi + lo) / 2;
    vector<long double> add, remove, all;
    int now = 0;
    for (int i = 0; i < n; i++) {
      long double d = norm(p[i]) / 2;
      if (d > mid) {
        continue;
      }
      long double a = sqrtl(mid * mid - d * d);
      long double t = atanl(a / d);
      long double f = atan2l(p[i].y, p[i].x);
      long double beg = f - t;
      long double end = f + t;
      if (beg < -pi) {
        beg += 2 * pi;
      }
      if (end > pi) {
        end -= 2 * pi;
      }
      if (end < beg) {
        now++;
      }
      add.emplace_back(beg);
      remove.emplace_back(end);
      all.emplace_back(beg);
      all.emplace_back(end);
    };
    sort(add.begin(), add.end());
    sort(remove.begin(), remove.end());
    sort(all.begin(), all.end());
    all.resize(unique(all.begin(), all.end()) - all.begin());
    int ok = 0;
    int it1 = 0, it2 = 0;
    if (now >= k) {
      ok = 1;
    }
    for (auto v : all) {
      while (it1 < (int)add.size() && abs(add[it1] - v) < eps) {
        now++;
        it1++;
      }
      while (it2 < (int)remove.size() && abs(remove[it2] - v) < eps) {
        now--;
        it2++;
      }
      if (now >= k) {
        ok = 1;
      }
    }
    if (ok) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  cout << fixed << setprecision(12);
  cout << hi << '\n';
  return 0;
}

```