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>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 200005;
struct dot {
  int x, y, id;
} a[N], b[N];
int sq[N];
double ans = 1e9;
int special, n, m, S, SS;
dot operator+(dot a, dot b) { return {a.x + b.x, a.y + b.y}; }
dot operator-(dot a, dot b) { return {a.x - b.x, a.y - b.y}; }
bool operator==(dot a, dot b) { return 1ll * a.x * b.y == 1ll * a.y * b.x; }
bool operator!=(dot a, dot b) { return 1ll * a.x * b.y != 1ll * a.y * b.x; }
inline double sqr(double x) { return x * x; }
inline double dist(int x, int y) {
  return sqrt(sqr(a[x].x - a[y].x) + sqr(a[x].y - a[y].y));
}
int main() {
  scanf("%d%d", &m, &SS);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &b[i].x, &b[i].y);
    b[i].id = i;
  }
  sort(b + 1, b + m + 1, [&](dot a, dot b) {
    return make_pair(a.x, a.y) <= make_pair(b.x, b.y);
  });
  for (int i = 1; i <= m; i++) {
    if (i == 1 ||
        make_pair(b[i - 1].x, b[i - 1].y) != make_pair(b[i].x, b[i].y))
      a[++n] = b[i];
    if (b[i].id == SS) S = n;
  }
  if (n == 1) {
    puts("0.000000");
    return 0;
  }
  if (n == 2) {
    printf("%.10f\n", dist(1, 2));
    return 0;
  }
  if (n == 3) {
    double ans = 1e9;
    for (int i = 1; i <= 3; i++) {
      for (int j = 1; j <= 3; j++) {
        if (i != S && j != S && i != j) {
          ckmin(ans, dist(S, i) + dist(i, j));
        }
      }
    }
    printf("%.10f\n", ans);
    return 0;
  }
  dot a1 = a[2] - a[1], a2 = a[3] - a[1], a3 = a[3] - a[2], v;
  if (a1 == a2)
    v = a1;
  else if (a1 == a2)
    v = a2;
  else
    v = a3, special = 1;
  if (!special) {
    for (int i = 2; i <= n; i++)
      if ((a[i] - a[1]) != v) special = i;
  }
  if (!special) special = 1;
  for (int i = 1; i <= n; i++) {
    if (i == S) {
      int t = 0;
      double sum = 0;
      if (i == special) {
        for (int j = 1; j <= n; j++) {
          if (j != special) sq[++t] = j;
        }
      } else {
        for (int j = i; j >= 1; j--)
          if (j != special) sq[++t] = j;
        for (int j = i + 1; j <= n; j++)
          if (j != special) sq[++t] = j;
      }
      for (int j = 1; j < n - 1; j++) sum += dist(sq[j], sq[j + 1]);
      if (i != special) {
        ckmin(ans, sum + dist(special, sq[n - 1]));
        for (int j = n - 2; j >= 1; j--)
          ckmin(ans, sum - dist(sq[j], sq[j + 1]) + dist(sq[j], special) +
                         dist(sq[j + 1], special));
      } else {
        ckmin(ans, sum + dist(special, sq[1]));
      }
      t = 0;
      if (i == special) {
        for (int j = n; j >= 1; j--)
          if (j != special) sq[++t] = j;
      } else {
        for (int j = i; j <= n; j++)
          if (j != special) sq[++t] = j;
        for (int j = i - 1; j >= 1; j--)
          if (j != special) sq[++t] = j;
      }
      sum = 0;
      for (int j = 1; j < n - 1; j++) sum += dist(sq[j], sq[j + 1]);
      if (i != special) {
        ckmin(ans, sum + dist(special, sq[n - 1]));
        for (int j = n - 2; j >= 1; j--)
          ckmin(ans, sum - dist(sq[j], sq[j + 1]) + dist(sq[j], special) +
                         dist(sq[j + 1], special));
      } else {
        ckmin(ans, sum + dist(special, sq[1]));
      }
    }
  }
  printf("%.10f\n", ans);
  return 0;
}
