#include <bits/stdc++.h>
using namespace std;
const int Inf = 0x3f3f3f3f;
int inline gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int N, K;
int a[200005], b[200005];
map<pair<int, int>, vector<int> > ps;
int id[200005];
double dist(int x, int y) {
  return sqrt(1ll * (a[x] - a[y]) * (a[x] - a[y]) +
              1ll * (b[x] - b[y]) * (b[x] - b[y]));
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; ++i) scanf("%d%d", a + i, b + i);
  for (int i = 2; i <= N; ++i) {
    int x = a[i] - a[1], y = b[i] - b[1];
    int g = gcd(x, y);
    x /= g;
    y /= g;
    if (y < 0 || (y == 0 && x < 0)) x *= -1, y *= -1;
    ps[make_pair(x, y)].push_back(i);
  }
  int kp;
  if (((int)(ps).size()) == 1)
    kp = 1;
  else if (((int)(ps).size()) == 2) {
    auto it = ps.begin();
    if (((int)(it->second).size()) == 1)
      kp = it->second[0];
    else
      kp = (++it)->second[0];
  } else
    kp = 1;
  int ids = 0;
  for (int i = 1; i <= N; ++i)
    if (kp != i) id[++ids] = i;
  sort(id + 1, id + N, [&](int x, int y) {
    return make_pair(a[x], b[x]) < make_pair(a[y], b[y]);
  });
  double ans;
  if (K == kp) {
    ans = dist(kp, id[1]) + dist(id[1], id[N - 1]);
    ans = min(ans, dist(kp, id[N - 1]) + dist(id[1], id[N - 1]));
  } else {
    int nid = 1;
    while (id[nid] != K) ++nid;
    ans = dist(K, id[1]) + dist(id[1], kp) +
          (nid == N - 1 ? 0
                        : dist(kp, id[nid + 1]) + dist(id[nid + 1], id[N - 1]));
    ans = min(
        ans,
        dist(K, id[N - 1]) + dist(id[N - 1], kp) +
            (nid == 1 ? 0 : dist(kp, id[nid - 1]) + dist(id[nid - 1], id[1])));
    ans = min(ans, dist(K, id[1]) + dist(id[1], kp) +
                       (nid == N - 1 ? 0
                                     : dist(kp, id[N - 1]) +
                                           dist(id[nid + 1], id[N - 1])));
    ans = min(ans,
              dist(K, id[N - 1]) + dist(id[N - 1], kp) +
                  (nid == 1 ? 0 : dist(kp, id[1]) + dist(id[nid - 1], id[1])));
    ans = min(ans, dist(K, kp) + dist(kp, id[1]) + dist(id[1], id[N - 1]));
    ans = min(ans, dist(K, kp) + dist(kp, id[N - 1]) + dist(id[1], id[N - 1]));
    ans =
        min(ans, dist(K, id[N - 1]) + dist(id[N - 1], id[1]) + dist(id[1], kp));
    ans =
        min(ans, dist(K, id[1]) + dist(id[N - 1], id[1]) + dist(id[N - 1], kp));
    for (int nk = 2; nk <= nid; ++nk)
      ans = min(ans, dist(K, id[N - 1]) + dist(id[N - 1], id[nk]) +
                         dist(id[nk], kp) + dist(kp, id[1]) +
                         dist(id[1], id[nk - 1]));
    for (int nk = nid; nk < N - 1; ++nk)
      ans = min(ans, dist(K, id[1]) + dist(id[1], id[nk]) + dist(id[nk], kp) +
                         dist(kp, id[N - 1]) + dist(id[N - 1], id[nk + 1]));
    for (int nk = 2; nk <= nid; ++nk)
      ans = min(ans, dist(K, id[N - 1]) + dist(id[N - 1], id[nk]) +
                         dist(id[nk], kp) + dist(kp, id[nk - 1]) +
                         dist(id[1], id[nk - 1]));
    for (int nk = nid; nk < N - 1; ++nk)
      ans = min(ans, dist(K, id[1]) + dist(id[1], id[nk]) + dist(id[nk], kp) +
                         dist(kp, id[nk + 1]) + dist(id[N - 1], id[nk + 1]));
  }
  printf("%.10lf\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int Inf = 0x3f3f3f3f;
int inline gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int N, K;
int a[200005], b[200005];
map<pair<int, int>, vector<int> > ps;
int id[200005];
double dist(int x, int y) {
  return sqrt(1ll * (a[x] - a[y]) * (a[x] - a[y]) +
              1ll * (b[x] - b[y]) * (b[x] - b[y]));
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; ++i) scanf("%d%d", a + i, b + i);
  for (int i = 2; i <= N; ++i) {
    int x = a[i] - a[1], y = b[i] - b[1];
    int g = gcd(x, y);
    x /= g;
    y /= g;
    if (y < 0 || (y == 0 && x < 0)) x *= -1, y *= -1;
    ps[make_pair(x, y)].push_back(i);
  }
  int kp;
  if (((int)(ps).size()) == 1)
    kp = 1;
  else if (((int)(ps).size()) == 2) {
    auto it = ps.begin();
    if (((int)(it->second).size()) == 1)
      kp = it->second[0];
    else
      kp = (++it)->second[0];
  } else
    kp = 1;
  int ids = 0;
  for (int i = 1; i <= N; ++i)
    if (kp != i) id[++ids] = i;
  sort(id + 1, id + N, [&](int x, int y) {
    return make_pair(a[x], b[x]) < make_pair(a[y], b[y]);
  });
  double ans;
  if (K == kp) {
    ans = dist(kp, id[1]) + dist(id[1], id[N - 1]);
    ans = min(ans, dist(kp, id[N - 1]) + dist(id[1], id[N - 1]));
  } else {
    int nid = 1;
    while (id[nid] != K) ++nid;
    ans = dist(K, id[1]) + dist(id[1], kp) +
          (nid == N - 1 ? 0
                        : dist(kp, id[nid + 1]) + dist(id[nid + 1], id[N - 1]));
    ans = min(
        ans,
        dist(K, id[N - 1]) + dist(id[N - 1], kp) +
            (nid == 1 ? 0 : dist(kp, id[nid - 1]) + dist(id[nid - 1], id[1])));
    ans = min(ans, dist(K, id[1]) + dist(id[1], kp) +
                       (nid == N - 1 ? 0
                                     : dist(kp, id[N - 1]) +
                                           dist(id[nid + 1], id[N - 1])));
    ans = min(ans,
              dist(K, id[N - 1]) + dist(id[N - 1], kp) +
                  (nid == 1 ? 0 : dist(kp, id[1]) + dist(id[nid - 1], id[1])));
    ans = min(ans, dist(K, kp) + dist(kp, id[1]) + dist(id[1], id[N - 1]));
    ans = min(ans, dist(K, kp) + dist(kp, id[N - 1]) + dist(id[1], id[N - 1]));
    ans =
        min(ans, dist(K, id[N - 1]) + dist(id[N - 1], id[1]) + dist(id[1], kp));
    ans =
        min(ans, dist(K, id[1]) + dist(id[N - 1], id[1]) + dist(id[N - 1], kp));
    for (int nk = 2; nk <= nid; ++nk)
      ans = min(ans, dist(K, id[N - 1]) + dist(id[N - 1], id[nk]) +
                         dist(id[nk], kp) + dist(kp, id[1]) +
                         dist(id[1], id[nk - 1]));
    for (int nk = nid; nk < N - 1; ++nk)
      ans = min(ans, dist(K, id[1]) + dist(id[1], id[nk]) + dist(id[nk], kp) +
                         dist(kp, id[N - 1]) + dist(id[N - 1], id[nk + 1]));
    for (int nk = 2; nk <= nid; ++nk)
      ans = min(ans, dist(K, id[N - 1]) + dist(id[N - 1], id[nk]) +
                         dist(id[nk], kp) + dist(kp, id[nk - 1]) +
                         dist(id[1], id[nk - 1]));
    for (int nk = nid; nk < N - 1; ++nk)
      ans = min(ans, dist(K, id[1]) + dist(id[1], id[nk]) + dist(id[nk], kp) +
                         dist(kp, id[nk + 1]) + dist(id[N - 1], id[nk + 1]));
  }
  printf("%.10lf\n", ans);
  return 0;
}

```