Problem: A new bus route is opened in the city <image>. The route is a closed polygon line in the place, with all segments parallel to one of the axes. m buses will operate on the route. All buses move in a loop along the route in the same direction with equal constant velocities (stopping times are negligible in this problem).

Buses start their movement in the first vertex of the route with equal interval. Suppose that T is the total time for a single bus to travel the whole loop of the route. Then, the bus 1 starts moving at time 0, the bus 2 starts at time T / m, the bus 3 starts at time 2T / m, and so on; finally, the bus m starts moving at time (m - 1)T / m. Thus, all intervals between pairs of consecutive buses (including the interval between the last and the first bus) are equal.

Buses can communicate with each other via wireless transmitters of equal power. If the transmitters have power D, then only buses within distance D of each other can communicate.

The buses are also equipped with a distributed system of schedule tracking. For all buses to stick to the schedule, the system has to synchronize the necessary data between all buses from time to time. At the moment of synchronization, the bus 1 communicates with the bus 2, the bus 2 — with bus 3, and so on; also, the bus m communicates with the bus 1.

As a research employee, you are tasked with finding the smallest value of D such that it is possible to find a time moment to perform synchronization once all buses have started moving. 

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 105) — the number of vertices of the polygonal line, and the number of buses respectively.

Next n lines describe the vertices of the route in the traversing order. Each of these lines contains two integers xi, yi ( - 1000 ≤ xi, yi ≤ 1000) — coordinates of respective vertex.

It is guaranteed that each leg of the route (including the leg between the last and the first vertex) is paralles to one of the coordinate axes. Moreover, no two subsequent vertices of the route coincide. The route is allowed to have self-intersections, and travel along the same segment multiple times.

Output

Print one real number — the answer to the problem. Your answer will be accepted if the relative or the absolute error doesn't exceed 10 - 6.

Examples

Input

4 2
0 0
0 1
1 1
1 0


Output

1.000000000


Input

2 2
0 0
1 0


Output

0.000000000

Note

Suppose that each bus travel 1 distance unit per second.

In the first sample case, in 0.5 seconds buses will be at distance 1, hence we can choose D = 1.

In the second sample case, in 0.5 seconds both buses will be at (0.5, 0), hence we can choose D = 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T read() {
  T f = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) f = c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  return f ? -x : x;
}
namespace run {
const int N = 4e5 + 9;
const double eps = 1e-8, inf = 1e9;
inline double squ(double x) { return x * x; }
struct node {
  double x, y;
  explicit node() {}
  node(double _x, double _y) : x(_x), y(_y) {}
  inline node operator+(const node &A) const { return node(x + A.x, y + A.y); }
  inline node operator-(const node &A) const { return node(x - A.x, y - A.y); }
  inline node operator*(const double &B) const { return node(x * B, y * B); }
} d[N];
inline void print(node A) { cout << A.x << " " << A.y << endl; }
inline double dist(node A, node B) {
  return sqrt(squ(A.x - B.x) + squ(A.y - B.y));
}
struct func1 {
  double k, b;
  explicit func1() {}
  func1(double _k, double _b) : k(_k), b(_b) {}
  inline func1 operator-(const func1 &A) const {
    return func1(k - A.k, b - A.b);
  }
};
inline func1 get(node A, node B) {
  double k = (A.y - B.y) / (A.x - B.x), b = A.y - k * A.x;
  return func1(k, b);
}
struct func2 {
  double l, r, a, b, c;
  explicit func2() {}
  func2(double _l, double _r, double _a, double _b, double _c)
      : l(_l), r(_r), a(_a), b(_b), c(_c) {}
} f[N];
inline func2 get(double l, double r, func1 x, func1 y) {
  double a = squ(x.k) + squ(y.k), b = 2. * x.k * x.b + 2. * y.k * y.b,
         c = squ(x.b) + squ(y.b);
  return func2(l, r, a, b, c);
}
int n, m, tot, top;
double S;
struct point {
  double pos;
  int v;
  explicit point() {}
  point(double _pos, int _v) : pos(_pos), v(_v) {}
  inline bool operator<(const point &A) const {
    if (fabs(pos - A.pos) > eps) return pos < A.pos;
    return v < A.v;
  }
} st[N << 1];
inline bool check(double mid) {
  top = 0;
  for (int i = 1; i <= tot; i++) {
    if (fabs(f[i].a) < eps) {
      double c = f[i].c - squ(mid);
      if (c < eps) {
        double x1 = f[i].l, x2 = f[i].r;
        int tmp = x1 / S;
        x1 -= S * tmp, x2 -= S * tmp;
        while (x2 > 0.) {
          st[++top] = point(x1, 1), st[++top] = point(x2, -1);
          x1 -= S, x1 = max(x1, (double)0.), x2 -= S;
        }
      }
      continue;
    }
    double c = f[i].c - squ(mid);
    double delta = f[i].b * f[i].b - 4. * f[i].a * c;
    if (delta < 0.) continue;
    delta = sqrt(delta);
    double x1 = (-f[i].b - delta) / (f[i].a * 2.),
           x2 = (-f[i].b + delta) / (f[i].a * 2.);
    x1 = max(x1 + f[i].l, f[i].l), x2 = min(x2 + f[i].l, f[i].r);
    if (x1 < x2) {
      int tmp = x1 / S;
      x1 -= S * tmp, x2 -= S * tmp;
      while (x2 > 0.) {
        st[++top] = point(x1, 1), st[++top] = point(x2, -1);
        x1 -= S, x1 = max(x1, (double)0.), x2 -= S;
      }
    }
  }
  sort(st + 1, st + top + 1);
  for (int i = 1, sum = 0; i <= top; i++) {
    sum += st[i].v;
    if (sum == m) return 1;
  }
  return 0;
}
inline double dich() {
  double l = 0., r = inf;
  for (int t = 100; t; t--) {
    double mid = 0.5 * (l + r);
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  return l;
}
inline int next(int id) { return id == n ? 1 : id + 1; }
int main() {
  n = read<int>(), m = read<int>();
  for (int i = 1; i <= n; i++) {
    int x = read<int>(), y = read<int>();
    d[i] = node(x, y);
    if (i > 1) S += dist(d[i], d[i - 1]);
  }
  S += dist(d[1], d[n]), S /= m;
  double sum = 0., tim = 0.;
  int L = 1, R = 1;
  node l = d[1], r;
  while (sum + dist(d[R], d[R + 1]) - eps <= S)
    sum += dist(d[R], d[R + 1]), R++;
  r = d[R], tim = 0.,
  r = r + (d[R + 1] - d[R]) * ((S - sum) / dist(d[R], d[R + 1]));
  do {
    double _l = dist(l, d[next(L)]), _r = dist(r, d[next(R)]);
    if (_l < _r) {
      func1 lx = get(node(0, l.x), node(_l, d[next(L)].x));
      func1 ly = get(node(0, l.y), node(_l, d[next(L)].y));
      double tmpx = r.x, tmpy = r.y;
      l = d[next(L)], L = next(L),
      r = r + (d[next(R)] - d[R]) * (_l / dist(d[next(R)], d[R]));
      func1 rx = get(node(0, tmpx), node(_l, r.x));
      func1 ry = get(node(0, tmpy), node(_l, r.y));
      if (_l > eps) {
        f[++tot] = get(tim, tim + _l, lx - rx, ly - ry);
      }
      tim += _l;
    } else {
      func1 rx = get(node(0, r.x), node(_r, d[next(R)].x));
      func1 ry = get(node(0, r.y), node(_r, d[next(R)].y));
      double tmpx = l.x, tmpy = l.y;
      r = d[next(R)], R = next(R),
      l = l + (d[next(L)] - d[L]) * (_r / dist(d[next(L)], d[L]));
      func1 lx = get(node(0, tmpx), node(_r, l.x));
      func1 ly = get(node(0, tmpy), node(_r, l.y));
      if (_r > eps) {
        f[++tot] = get(tim, tim + _r, lx - rx, ly - ry);
      }
      tim += _r;
    }
  } while (tim <= S * m);
  printf("%.10lf\n", dich());
  return 0;
}
}  // namespace run
int main() { return run::main(); }