#include <bits/stdc++.h>
using namespace std;
double x[200005], y[200005], k, b, Dn[200005], Dn1[200005], D1[200005];
int XX[200005], YY[200005];
pair<double, double> B[200005];
pair<double, int> a[200005];
int n;
inline int check(int i, int j) {
  k = (y[j] - y[i]) / (x[j] - x[i]);
  b = y[i] - k * x[i];
  int pos = -1;
  for (int i = 1; i <= n; i++) {
    if (fabs(y[i] - x[i] * k - b) >= 1e-8) {
      if (pos == -1)
        pos = i;
      else
        return -1;
    }
  }
  return pos;
}
inline double dis(int i, int j) {
  double X = x[a[i].second] - x[a[j].second],
         Y = y[a[i].second] - y[a[j].second];
  return sqrt(X * X + Y * Y);
}
map<pair<double, double>, int> mp;
int main(int argc, char** argv) {
  int s, p;
  ios::sync_with_stdio(false);
  cin.tie(0);
  srand((unsigned long long)new char);
  cin >> n >> s;
  for (int i = 1; i <= n; i++) {
    cin >> XX[i] >> YY[i];
    x[i] = XX[i], y[i] = YY[i];
    B[i] = {x[i], y[i]};
  }
  pair<double, double> S = B[s];
  sort(B + 1, B + n + 1);
  int cnt = 0;
  x[0] = 1e9;
  for (int i = 1; i <= n; i++)
    if (B[i] != make_pair(x[cnt], y[cnt]))
      x[++cnt] = B[i].first, y[cnt] = B[i].second;
  for (int i = 1; i <= cnt; i++)
    if (make_pair(x[i], y[i]) == S) s = i;
  n = cnt;
  int cnt1 = 0, cnt2 = 0;
  for (int i = 1; i <= n; i++) {
    if (x[i] == x[1])
      ++cnt1;
    else if (x[i] == x[2])
      ++cnt2;
  }
  if (cnt1 == n - 1) {
    for (int i = 1; i <= n; i++)
      if (x[i] != x[1]) p = i;
  } else if (cnt2 == n - 1) {
    for (int i = 1; i <= n; i++)
      if (x[i] != x[2]) p = i;
  } else if ((p = check(1, 2)) == -1)
    if ((p = check(3, 2)) == -1) p = check(1, 3);
  double X = (double)rand() / rand(), Y = (double)rand() / rand();
  while (X < 0.1) X = (double)rand() / (double)rand();
  while (Y < 0.1) Y = (double)rand() / (double)rand();
  cnt = 0;
  for (int i = 1; i <= n; i++)
    if (i != p) a[++cnt].first = X * x[i] + Y * y[i], a[cnt].second = i;
  sort(a + 1, a + n);
  a[n].second = p;
  double ans = 1e100;
  if (p == s) {
    ans = min(ans, dis(n, 1) + dis(1, n - 1));
    ans = min(ans, dis(n, n - 1) + dis(1, n - 1));
    printf("%.10lf", ans);
  } else {
    for (int i = 1; i <= n; i++)
      D1[i] = dis(1, i), Dn[i] = dis(n, i), Dn1[i] = dis(n - 1, i);
    for (int i = 1; i < n; i++) {
      if (a[i].second == s) {
        ans = min(dis(i, n - 1) + dis(n - 1, n) +
                      (i == 1 ? 0 : dis(n, i - 1) + dis(i - 1, 1)),
                  dis(i, 1) + dis(1, n) +
                      (i == n - 1 ? 0 : dis(n - 1, i + 1) + dis(n, i + 1)));
        for (int j = 1; j < n - 1; j++) {
          ans = min(ans, D1[i] + D1[j] + Dn[j] + Dn[j + 1] + Dn1[j + 1]);
          ans = min(ans, Dn1[i] + Dn1[j + 1] + Dn[j + 1] + Dn[j] + D1[j]);
        }
        ans = min(ans, dis(i, 1) + dis(1, n - 1) + dis(n - 1, n));
        ans = min(ans, dis(i, n - 1) + dis(1, n - 1) + dis(n, 1));
        break;
      }
    }
    printf("%.10lf", 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;
double x[200005], y[200005], k, b, Dn[200005], Dn1[200005], D1[200005];
int XX[200005], YY[200005];
pair<double, double> B[200005];
pair<double, int> a[200005];
int n;
inline int check(int i, int j) {
  k = (y[j] - y[i]) / (x[j] - x[i]);
  b = y[i] - k * x[i];
  int pos = -1;
  for (int i = 1; i <= n; i++) {
    if (fabs(y[i] - x[i] * k - b) >= 1e-8) {
      if (pos == -1)
        pos = i;
      else
        return -1;
    }
  }
  return pos;
}
inline double dis(int i, int j) {
  double X = x[a[i].second] - x[a[j].second],
         Y = y[a[i].second] - y[a[j].second];
  return sqrt(X * X + Y * Y);
}
map<pair<double, double>, int> mp;
int main(int argc, char** argv) {
  int s, p;
  ios::sync_with_stdio(false);
  cin.tie(0);
  srand((unsigned long long)new char);
  cin >> n >> s;
  for (int i = 1; i <= n; i++) {
    cin >> XX[i] >> YY[i];
    x[i] = XX[i], y[i] = YY[i];
    B[i] = {x[i], y[i]};
  }
  pair<double, double> S = B[s];
  sort(B + 1, B + n + 1);
  int cnt = 0;
  x[0] = 1e9;
  for (int i = 1; i <= n; i++)
    if (B[i] != make_pair(x[cnt], y[cnt]))
      x[++cnt] = B[i].first, y[cnt] = B[i].second;
  for (int i = 1; i <= cnt; i++)
    if (make_pair(x[i], y[i]) == S) s = i;
  n = cnt;
  int cnt1 = 0, cnt2 = 0;
  for (int i = 1; i <= n; i++) {
    if (x[i] == x[1])
      ++cnt1;
    else if (x[i] == x[2])
      ++cnt2;
  }
  if (cnt1 == n - 1) {
    for (int i = 1; i <= n; i++)
      if (x[i] != x[1]) p = i;
  } else if (cnt2 == n - 1) {
    for (int i = 1; i <= n; i++)
      if (x[i] != x[2]) p = i;
  } else if ((p = check(1, 2)) == -1)
    if ((p = check(3, 2)) == -1) p = check(1, 3);
  double X = (double)rand() / rand(), Y = (double)rand() / rand();
  while (X < 0.1) X = (double)rand() / (double)rand();
  while (Y < 0.1) Y = (double)rand() / (double)rand();
  cnt = 0;
  for (int i = 1; i <= n; i++)
    if (i != p) a[++cnt].first = X * x[i] + Y * y[i], a[cnt].second = i;
  sort(a + 1, a + n);
  a[n].second = p;
  double ans = 1e100;
  if (p == s) {
    ans = min(ans, dis(n, 1) + dis(1, n - 1));
    ans = min(ans, dis(n, n - 1) + dis(1, n - 1));
    printf("%.10lf", ans);
  } else {
    for (int i = 1; i <= n; i++)
      D1[i] = dis(1, i), Dn[i] = dis(n, i), Dn1[i] = dis(n - 1, i);
    for (int i = 1; i < n; i++) {
      if (a[i].second == s) {
        ans = min(dis(i, n - 1) + dis(n - 1, n) +
                      (i == 1 ? 0 : dis(n, i - 1) + dis(i - 1, 1)),
                  dis(i, 1) + dis(1, n) +
                      (i == n - 1 ? 0 : dis(n - 1, i + 1) + dis(n, i + 1)));
        for (int j = 1; j < n - 1; j++) {
          ans = min(ans, D1[i] + D1[j] + Dn[j] + Dn[j + 1] + Dn1[j + 1]);
          ans = min(ans, Dn1[i] + Dn1[j + 1] + Dn[j + 1] + Dn[j] + D1[j]);
        }
        ans = min(ans, dis(i, 1) + dis(1, n - 1) + dis(n - 1, n));
        ans = min(ans, dis(i, n - 1) + dis(1, n - 1) + dis(n, 1));
        break;
      }
    }
    printf("%.10lf", ans);
  }
  return 0;
}

```