#include <bits/stdc++.h>
using namespace std;
class Point {
 public:
  double x, y;
  Point() {}
  Point(double x, double y) {
    (*this).x = x;
    (*this).y = y;
  }
  double operator^(const Point &p) const { return x * p.y - y * p.x; }
  double operator*(const Point &p) const { return x * p.x + y * p.y; }
  Point operator*(const double &d) const { return Point(x * d, y * d); }
  Point operator/(const double &d) const { return Point(x / d, y / d); }
  Point operator-(const Point &p) const { return Point(x - p.x, y - p.y); }
  Point operator+(const Point &p) const { return Point(x + p.x, y + p.y); }
  double sqr() { return x * x + y * y; }
  double abs() { return sqrt(sqr()); }
  double distance(const Point &p) { return fabs((*this - p).abs()); }
  void print() { printf("%.8lf %.8lf", x, y); }
};
class Line {
 public:
  Point p1, p2;
  Point vec;
  Line() {}
  Line(Point p1, Point p2) {
    (*this).p1 = p1;
    (*this).p2 = p2;
    (*this).vec = p2 - p1;
  }
};
class Circle {
 public:
  Point o;
  double r;
  Circle() {}
  Circle(Point o, double r) {
    (*this).o = o;
    (*this).r = r;
  }
  Circle(double cx, double cy, double r) {
    (*this).o.x = cx;
    (*this).o.y = cy;
    (*this).r = r;
  }
};
Point polar(double length, double angle) {
  return Point(length * cos(angle), length * sin(angle));
}
pair<Point, Point> get_Cross_Points(Circle c1, Circle c2) {
  Point vec12 = c2.o - c1.o;
  double d = vec12.abs();
  double a = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2.0 * c1.r * d));
  double t = atan2(vec12.y, vec12.x);
  return make_pair(c1.o + polar(c1.r, t - a), c1.o + polar(c1.r, t + a));
}
const double pi = 3.14159265358979;
double angle(double x, double y) {
  if (x == 0 && y > 0) return pi / 2;
  if (x == 0 && y < 0) return 3 * pi / 2;
  if (x > 0 && y == 0) return 0;
  if (x < 0 && y == 0) return pi;
  if (x > 0 && y > 0) return atan(y / x);
  if (x < 0 && y > 0) return pi + atan(y / x);
  if (x < 0 && y < 0) return pi + atan(y / x);
  if (x > 0 && y < 0) return pi * 2 + atan(y / x);
}
inline pair<double, double> qwq(Circle c1, Circle c2) {
  if (c1.o.x == 0 && c1.o.y == 0) return {-10000, 10000};
  if (c1.o.x * c1.o.x + c1.o.y * c1.o.y + 1E-8 > c1.r * c1.r * 4)
    return {1, -1145141919};
  pair<Point, Point> a = get_Cross_Points(c1, c2);
  double x = angle(a.first.x, a.first.y);
  double y = angle(a.second.x, a.second.y);
  if (x > y) swap(x, y);
  if (y - x >= pi) x += pi * 2, swap(x, y);
  return {x, y};
}
Point a[100005];
Circle b[100005];
int n, k;
pair<double, double> c[200005];
priority_queue<double, vector<double>, greater<double> > q;
inline bool check(double x) {
  for (int i = 1; i <= n; i++) b[i].o = a[i], b[i].r = x;
  b[0].r = x, b[0].o = {0, 0};
  for (int i = 1; i <= n; i++) {
    c[i] = qwq(b[i], b[0]);
    if (c[i].second < c[i].first) c[i].second += pi * 2;
    c[i + n] = {c[i].first + pi * 2, c[i].second + pi * 2};
  }
  while (!q.empty()) q.pop();
  sort(c + 1, c + n + n + 1);
  for (int i = 1; i <= n + n; i++) {
    q.push(c[i].second);
    while (!q.empty() && q.top() < c[i].first) q.pop();
    if (q.size() >= k) return 1;
  }
  return 0;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
  double l = 0, r = 200000;
  while (l + (1e-6) <= r) {
    double mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%.10lf", (l + r) / 2);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
class Point {
 public:
  double x, y;
  Point() {}
  Point(double x, double y) {
    (*this).x = x;
    (*this).y = y;
  }
  double operator^(const Point &p) const { return x * p.y - y * p.x; }
  double operator*(const Point &p) const { return x * p.x + y * p.y; }
  Point operator*(const double &d) const { return Point(x * d, y * d); }
  Point operator/(const double &d) const { return Point(x / d, y / d); }
  Point operator-(const Point &p) const { return Point(x - p.x, y - p.y); }
  Point operator+(const Point &p) const { return Point(x + p.x, y + p.y); }
  double sqr() { return x * x + y * y; }
  double abs() { return sqrt(sqr()); }
  double distance(const Point &p) { return fabs((*this - p).abs()); }
  void print() { printf("%.8lf %.8lf", x, y); }
};
class Line {
 public:
  Point p1, p2;
  Point vec;
  Line() {}
  Line(Point p1, Point p2) {
    (*this).p1 = p1;
    (*this).p2 = p2;
    (*this).vec = p2 - p1;
  }
};
class Circle {
 public:
  Point o;
  double r;
  Circle() {}
  Circle(Point o, double r) {
    (*this).o = o;
    (*this).r = r;
  }
  Circle(double cx, double cy, double r) {
    (*this).o.x = cx;
    (*this).o.y = cy;
    (*this).r = r;
  }
};
Point polar(double length, double angle) {
  return Point(length * cos(angle), length * sin(angle));
}
pair<Point, Point> get_Cross_Points(Circle c1, Circle c2) {
  Point vec12 = c2.o - c1.o;
  double d = vec12.abs();
  double a = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2.0 * c1.r * d));
  double t = atan2(vec12.y, vec12.x);
  return make_pair(c1.o + polar(c1.r, t - a), c1.o + polar(c1.r, t + a));
}
const double pi = 3.14159265358979;
double angle(double x, double y) {
  if (x == 0 && y > 0) return pi / 2;
  if (x == 0 && y < 0) return 3 * pi / 2;
  if (x > 0 && y == 0) return 0;
  if (x < 0 && y == 0) return pi;
  if (x > 0 && y > 0) return atan(y / x);
  if (x < 0 && y > 0) return pi + atan(y / x);
  if (x < 0 && y < 0) return pi + atan(y / x);
  if (x > 0 && y < 0) return pi * 2 + atan(y / x);
}
inline pair<double, double> qwq(Circle c1, Circle c2) {
  if (c1.o.x == 0 && c1.o.y == 0) return {-10000, 10000};
  if (c1.o.x * c1.o.x + c1.o.y * c1.o.y + 1E-8 > c1.r * c1.r * 4)
    return {1, -1145141919};
  pair<Point, Point> a = get_Cross_Points(c1, c2);
  double x = angle(a.first.x, a.first.y);
  double y = angle(a.second.x, a.second.y);
  if (x > y) swap(x, y);
  if (y - x >= pi) x += pi * 2, swap(x, y);
  return {x, y};
}
Point a[100005];
Circle b[100005];
int n, k;
pair<double, double> c[200005];
priority_queue<double, vector<double>, greater<double> > q;
inline bool check(double x) {
  for (int i = 1; i <= n; i++) b[i].o = a[i], b[i].r = x;
  b[0].r = x, b[0].o = {0, 0};
  for (int i = 1; i <= n; i++) {
    c[i] = qwq(b[i], b[0]);
    if (c[i].second < c[i].first) c[i].second += pi * 2;
    c[i + n] = {c[i].first + pi * 2, c[i].second + pi * 2};
  }
  while (!q.empty()) q.pop();
  sort(c + 1, c + n + n + 1);
  for (int i = 1; i <= n + n; i++) {
    q.push(c[i].second);
    while (!q.empty() && q.top() < c[i].first) q.pop();
    if (q.size() >= k) return 1;
  }
  return 0;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
  double l = 0, r = 200000;
  while (l + (1e-6) <= r) {
    double mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%.10lf", (l + r) / 2);
}

```