You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and the required number of equal elements.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
double argminimal(function<double(double)> f, double l, double r,
                  int iter = 80) {
  for (int i = 0; i < iter; ++i) {
    double c1 = (l * 2 + r) / 3, c2 = (l + r * 2) / 3;
    if (f(c1) > f(c2)) {
      l = c1;
    } else {
      r = c2;
    }
  }
  return l;
}
double minimal(function<double(double)> f, double l, double r, int iter = 80) {
  return f(argminimal(f, l, r, iter));
}
double argmaximal(function<double(double)> f, double l, double r,
                  int iter = 80) {
  for (int i = 0; i < iter; ++i) {
    double c1 = (l * 2 + r) / 3, c2 = (l + r * 2) / 3;
    if (f(c1) > f(c2)) {
      r = c2;
    } else {
      l = c1;
    }
  }
  return l;
}
double maximul(function<double(double)> f, double l, double r, int iter = 80) {
  return f(argmaximal(f, l, r, iter));
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<long long> a(n);
  map<int, long long> hist;
  for (int i = 0; i < (int)(n); ++i) {
    scanf("%lld", &a[i]);
    ++hist[a[i]];
  }
  for (auto &p : hist) {
    if (p.second >= k) {
      cout << 0 << endl;
      return 0;
    }
  }
  long long ans = 1e18;
  {
    auto achievable = [&](int m) {
      int c = 0;
      for (int i = 0; i < (int)(n); ++i) c += (a[i] >= m);
      return c >= k;
    };
    int l = 0, r = 1e9 + 1;
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (achievable(m)) {
        l = m;
      } else {
        r = m;
      }
    }
    long long res = 0;
    for (int i = 0; i < (int)(n); ++i) res += max(a[i] - (l + 1), 0LL);
    res += k - hist[l];
    chmin(ans, res);
  }
  {
    auto achievable = [&](int m) {
      int c = 0;
      for (int i = 0; i < (int)(n); ++i) c += (a[i] <= m);
      return c >= k;
    };
    int l = 0, r = 1e9;
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (achievable(m)) {
        r = m;
      } else {
        l = m;
      }
    }
    long long res = 0;
    for (int i = 0; i < (int)(n); ++i) res += max(r - 1 - a[i], 0LL);
    res += k - hist[r];
    chmin(ans, res);
  }
  {
    auto cost = [&](int m) {
      long long c = 0;
      for (int i = 0; i < (int)(n); ++i) {
        if (a[i] == m) continue;
        c += max(a[i] - (m + 1), 0LL);
        c += max(m - 1 - a[i], 0LL);
      }
      c += k - hist[m];
      return c;
    };
    auto negativeGradient = [&](int m) { return cost(m + 1) - cost(m) < 0; };
    int l = 0, r = 1e9;
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (negativeGradient(m)) {
        l = m;
      } else {
        r = m;
      }
    }
    chmin(ans, cost(r));
  }
  cout << ans << endl;
  return 0;
}
