Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void error() { cerr << endl; }
template <typename Head, typename... Tail>
void error(Head H, Tail... T) {
  cerr << ' ' << H;
  error(T...);
}
void Solve();
int main() {
  Solve();
  return 0;
}
using LL = long long;
const int N = 2e6 + 5, INF = 0x3f3f3f3f;
int a[N];
LL sum[N];
LL ans = LONG_LONG_MAX;
void Solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (int i = m; i <= n; i++)
    if (a[i] == a[i - m + 1]) ans = 0;
  if (ans == 0) return cout << 0 << endl, void();
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
  for (int i = 1; i <= n; i++) {
    if (i >= m) ans = min(ans, (LL)i * a[i] - sum[i] - (i - m));
  }
  error(ans);
  for (int i = n; i >= 1; i--) {
    if (n - i + 1 >= m)
      ans = min(ans, sum[n] - sum[i - 1] - (LL)(n - i + 1) * a[i] -
                         ((n - i + 1) - m));
  }
  for (int i = 1; i <= n; i++) {
    ans = min(ans, (LL)i * a[i] - sum[i] + sum[n] - sum[i - 1] -
                       (LL)(n - i + 1) * a[i] - (n - m));
  }
  cout << ans << endl;
}