Problem: Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.

Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.

Scrooge signed exactly k papers throughout his life and all those signatures look the same.

Find the total time Scrooge wasted signing the papers.

Input

The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.

All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.

Output

Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.

Examples

Input

2 1
0 0
10 0


Output

0.200000000

Input

5 10
3 1
-5 6
-2 -1
3 2
10 0


Output

6.032163204

Input

6 10
5 0
4 0
6 0
3 0
7 0
2 0


Output

3.000000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1 << 29;
const long long MOD = 1000000007;
const long long FMOD = 998244353;
const long long MAX = 10000000005;
const long long MIN = -10000000005;
int cmp(const void* a, const void* b);
long long maxv(long long a, long long b);
long long minv(long long a, long long b);
long long gcd(long long u, long long v);
long long digits(long long n);
bool ispoweroftwo(long long n);
bool isvowel(char x);
long long chartoint(char ch);
long long CEIL(long long x, long long y);
long long FLOOR(long long x, long long y);
int main() {
  long long n, k;
  scanf("%lld%lld", &n, &k);
  double ans = 0;
  long long px, py;
  scanf("%lld%lld", &px, &py);
  for (long long i = 0; i < (n - 1); ++i) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    double z = sqrt((x - px) * (x - px) + (y - py) * (y - py));
    ans += z;
    px = x;
    py = y;
  }
  ans = ans * k / 50.0;
  cout << fixed << setprecision(10) << ans << "\n";
  return 0;
}
int cmp(const void* a, const void* b) {
  if (*(long long*)a - *(long long*)b < 0) return -1;
  if (*(long long*)a - *(long long*)b > 0) return 1;
  return 0;
}
long long maxv(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long minv(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long gcd(long long u, long long v) {
  if (v == 0) return u;
  return gcd(v, u % v);
}
long long digits(long long n) { return floor(log10(n)) + 1; }
bool ispoweroftwo(long long x) { return x && (!(x & (x - 1))); }
bool isvowel(char x) {
  return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u');
}
long long chartoint(char ch) {
  if (ch >= 'A' && ch <= 'Z')
    return (ch - 'A');
  else if (ch >= '0' && ch <= '9')
    return (ch - '0');
  else if (ch >= 'a' && ch <= 'z')
    return (ch - 'a');
  else
    return 0;
}
long long CEIL(long long x, long long y) {
  if (x % y == 0)
    return (x / y);
  else
    return (x / y + 1);
}
long long FLOOR(long long x, long long y) {
  if (x % y == 0)
    return (x / y);
  else
    return (x / y - 1);
}