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 std::cin;
using std::cout;
const int maxn = 200200;
struct pt {
  long double x, y;
};
int x[maxn], y[maxn];
pt node[maxn];
int n, m;
long double T, S;
inline pt sub(const pt& x, const pt& y) { return (pt){x.x - y.x, x.y - y.y}; }
inline pt add(const pt& x, const pt& y) { return (pt){x.x + y.x, x.y + y.y}; }
inline pt mul(const pt& x, long double y) { return (pt){x.x * y, x.y * y}; }
inline long double sqr(long double x) { return x * x; }
inline long double dis(const pt& x) { return sqrt(abs(x.x) + abs(x.y)); }
inline long double dis2(const pt& x) { return abs(x.x) + abs(x.y); }
inline long double dis(const pt& x, const pt& y) {
  return sqrt(sqr(x.x - y.x) + sqr(x.y - y.y));
}
inline long double dis2(const pt& x, const pt& y) {
  return sqr(x.x - y.x) + sqr(x.y - y.y);
}
std::vector<std::pair<long double, long double> > vc;
struct func {
  long double a, b, c;
};
static const long double eps = 1e-12, inf = 1. / 0;
inline std::pair<long double, long double> getroot(func fc) {
  static const std::pair<long double, long double> nosol = {-inf, inf};
  static const std::pair<long double, long double> allsol = {inf, -inf};
  if (fc.a < eps) return fc.c <= 0 ? nosol : allsol;
  const long double delta = fc.b * fc.b - 4 * fc.a * fc.c;
  if (delta < 0) return allsol;
  const long double x = sqrt(delta);
  return {
      (-fc.b - x) / fc.a / 2,
      (-fc.b + x) / fc.a / 2,
  };
}
inline void push(long double l, long double r) {
  l = fmod(l, T);
  r = fmod(r, T);
  if (r < l) {
    if (r < 0 || l > T) exit(1);
    vc.emplace_back(0, r);
    vc.emplace_back(l, T);
  } else {
    if (l > r || l < 0 || r > T) {
      exit(2);
    }
    vc.emplace_back(l, r);
  }
}
inline void push(pt a0, pt b0, pt a1, pt b1, long double x, long double su) {
  const long double d = dis(a0, b0);
  if (d < eps) return;
  a0 = sub(a0, a1), b0 = sub(b0, b1);
  pt dt = mul(sub(b0, a0), 1. / d);
  std::pair<long double, long double> T =
      getroot((func){dt.x * dt.x + dt.y * dt.y, 2 * (a0.x * dt.x + a0.y * dt.y),
                     a0.x * a0.x + a0.y * a0.y - x * x});
  if (0 <= T.first) push(su, su + std::min<long double>(d, T.first));
  if (T.second <= d) push(su + std::max<long double>(0, T.second), su + d);
}
inline long double can(long double x) {
  pt x0 = node[0];
  int id0 = 1;
  pt x1 = node[0];
  int id1 = 1;
  long double rem = T;
  for (; rem >= dis(x1, node[id1]); ++id1 %= n)
    rem -= dis(x1, node[id1]), x1 = node[id1];
  x1 = add(x1, mul(sub(node[id1], x1), rem / dis(x1, node[id1])));
  long double su = 0;
  vc.clear();
  for (int i = 0; i < n + n; ++i) {
    if (dis(x0, node[id0]) > dis(x1, node[id1]))
      std::swap(id0, id1), std::swap(x0, x1);
    const long double d = dis(x0, node[id0]);
    const pt t0 = node[id0];
    const pt t1 = add(x1, mul(sub(node[id1], x1), d / dis(x1, node[id1])));
    push(x0, t0, x1, t1, x, su);
    su += d;
    if (su >= T) su -= T;
    x0 = t0;
    if (++id0 == n) id0 = 0;
    x1 = t1;
  }
  sort(vc.begin(), vc.end());
  long double max = 0, sum = 0;
  for (const std::pair<long double, long double>& i : vc) {
    if (max < i.first) {
      sum += i.second - i.first;
      max = i.second;
    } else {
      if (i.second > max) {
        sum += i.second - max;
        max = i.second;
      }
    }
  }
  return std::fabs(sum - T) > eps;
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i];
    node[i] = (pt){(long double)x[i], (long double)y[i]};
  }
  for (int i = 0; i < n; ++i) {
    S += dis(node[i], node[(i + 1) % n]);
  }
  T = S / m;
  long double l = 0, r = 2e4;
  for (int c = 0; c < 50; ++c) {
    long double mid = (l + r) / 2;
    if (can(mid)) {
      r = mid;
    } else {
      l = mid;
    }
  }
  printf("%.10lf\n", double(r));
}