You are given N points on an infinite plane with the Cartesian coordinate system on it. N-1 points lay on one line, and one point isn't on that line. You are on point K at the start, and the goal is to visit every point. You can move between any two points in a straight line, and you can revisit points. What is the minimum length of the path? 

Input

The first line contains two integers: N (3 ≤ N ≤ 2*10^5) - the number of points, and K (1 ≤ K ≤ N) - the index of the starting point.

Each of the next N lines contain two integers, A_i, B_i (-10^6 ≤ A_i, B_i ≤ 10^6) - coordinates of the i-th point.

Output

The output contains one number - the shortest path to visit all given points starting from point K. The absolute difference between your solution and the main solution shouldn't exceed 10^-6;

Example

Input


5 2
0 0
-1 1
2 -2
0 1
-2 2


Output


7.478709

Note

The shortest path consists of these moves: 

2 -> 5 

5 -> 4 

4 -> 1 

1 -> 3 

There isn't any shorter path possible.
4 1
768074 -27
-768090 -27
-8 -27
595283 233253
 4 3
-733 -804
-4 -755
725 -706
-895 -822
 16 8
-562856 268
-422202 268
281068 268
703030 268
-984818 268
-240 268
-844164 268
421722 268
-281548 268
34672 722872
562376 268
-703510 268
-140894 268
140414 268
843684 268
984338 268

1697753.3898821024 1624.2867736884 3781378.4984290055
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
struct Vector {
  int x, y;
  Vector(void) {}
  Vector(int _x, int _y) : x(_x), y(_y) {}
};
inline bool operator==(const Vector &p, const Vector &q) {
  return p.x == q.x && p.y == q.y;
}
inline Vector operator-(const Vector &p, const Vector &q) {
  return Vector(p.x - q.x, p.y - q.y);
}
inline long long cross(const Vector &p, const Vector &q) {
  return (long long)p.x * q.y - (long long)p.y * q.x;
}
inline long double dist(const Vector &p, const Vector &q) {
  return sqrtl((long long)(p.x - q.x) * (p.x - q.x) +
               (long long)(p.y - q.y) * (p.y - q.y));
}
inline bool coline(const Vector &p1, const Vector &p2, const Vector &p3) {
  return cross(p2 - p1, p3 - p1) == 0;
}
Vector p[MAXN];
int main(void) {
  int n, d;
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &p[i].x, &p[i].y);
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  if (n == 2) {
    printf("%.20lf\n", (double)dist(p[1], p[2]));
    return 0;
  }
  if (n == 3) {
    swap(p[d], p[1]);
    long double ans1 = dist(p[1], p[2]) + dist(p[2], p[3]);
    long double ans2 = dist(p[1], p[3]) + dist(p[3], p[2]);
    printf("%.20lf\n", (double)min(ans1, ans2));
    return 0;
  }
  Vector v1, v2;
  if (coline(p[1], p[2], p[3]))
    v1 = p[1], v2 = p[2];
  else if (coline(p[2], p[3], p[4]))
    v1 = p[2], v2 = p[3];
  else if (coline(p[1], p[3], p[4]))
    v1 = p[1], v2 = p[3];
  else
    v1 = p[1], v2 = p[2];
  int sp = d;
  for (int i = 1; i <= n; ++i)
    if (!coline(v1, v2, p[i])) {
      sp = i;
      break;
    }
  swap(p[sp], p[n]);
  Vector pd = p[d];
  sort(p + 1, p + n, [](Vector pp, Vector qq) {
    return pp.x == qq.x ? pp.y < qq.y : pp.x < qq.x;
  });
  d = find(p + 1, p + n + 1, pd) - p;
  if (d == n) {
    long double ans =
        min(dist(p[n], p[1]), dist(p[n], p[n - 1])) + dist(p[n - 1], p[1]);
    printf("%.20lf\n", (double)ans);
    return 0;
  }
  auto cmp_dis = [&](Vector pp, Vector qq) {
    return dist(pp, p[n]) < dist(qq, p[n]);
  };
  int mid1 = min_element(p + 1, p + n, cmp_dis) - p;
  int mid2;
  if (mid1 == 1)
    mid2 = mid1 + 1;
  else if (mid1 == n - 1)
    mid2 = mid1 - 1;
  else
    mid2 = cmp_dis(p[mid1 - 1], p[mid1 + 1]) ? mid1 - 1 : mid1 + 1;
  vector<int> vec = {1, n - 1, mid1, mid2, d};
  if (d > 1) vec.push_back(d - 1);
  if (d < n - 1) vec.push_back(d + 1);
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  vector<pair<int, int> > must;
  for (int i = 0; i + 1 < (int)vec.size(); ++i)
    if (vec[i + 1] - vec[i] != 1) must.emplace_back(vec[i], vec[i + 1]);
  vec.push_back(n);
  long double ans = 1e18;
  do {
    if (vec[0] != d) continue;
    static int pos[MAXN];
    for (int i = 0; i < (int)vec.size(); ++i) pos[vec[i]] = i;
    bool ok = 1;
    for (pair<int, int> cur : must)
      if (abs(pos[cur.first] - pos[cur.second]) != 1) {
        ok = 0;
        break;
      }
    if (!ok) continue;
    long double cur = 0;
    for (int i = 0; i + 1 < (int)vec.size(); ++i)
      cur += dist(p[vec[i]], p[vec[i + 1]]);
    ans = min(ans, cur);
  } while (next_permutation(vec.begin(), vec.end()));
  printf("%.20lf", (double)ans);
  return 0;
}
