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>
using namespace std;
char _buf[100000], *_op(_buf), *_ed(_buf);
inline int getint() {
  int _s = 0, _f = 1;
  char _ch = (_op == _ed && (_ed = (_op = _buf) + fread(_buf, 1, 100000, stdin),
                             _op == _ed)
                  ? EOF
                  : *_op++);
  while (!('0' <= _ch && _ch <= '9'))
    (_ch == '-') && (_f = -1),
        _ch =
            (_op == _ed && (_ed = (_op = _buf) + fread(_buf, 1, 100000, stdin),
                            _op == _ed)
                 ? EOF
                 : *_op++);
  while (('0' <= _ch && _ch <= '9'))
    _s = _s * 10 + _ch - 48,
    _ch = (_op == _ed && (_ed = (_op = _buf) + fread(_buf, 1, 100000, stdin),
                          _op == _ed)
               ? EOF
               : *_op++);
  return _s * _f;
}
const int N = 2e5 + 2;
int n, k, a[N];
long long int pre[N], suf[N], ans = 1ll << 62;
inline void input() {
  n = getint(), k = getint();
  for (int i = 1; i <= n; i++) a[i] = getint();
}
inline void init() { sort(a + 1, a + 1 + n); }
inline void work() {
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + a[i];
  for (int l = 1, r; l <= n; l = r + 1) {
    for (r = l; a[r + 1] == a[l]; r++)
      ;
    if (r - l + 1 >= k) return void(cout << 0);
    long long int tmp = 0;
    tmp += 1ll * (a[l] - 1) * (l - 1) - pre[l - 1];
    tmp += suf[r + 1] - 1ll * (a[r] + 1) * (n - r);
    tmp += k - (r - l + 1);
    if (ans > tmp) ans = tmp;
    if (r >= k) {
      tmp = 1ll * (a[l] - 1) * (l - 1) - pre[l - 1] + k - (r - l + 1);
      if (ans > tmp) ans = tmp;
    }
    if (n - l + 1 >= k) {
      tmp = suf[r + 1] - 1ll * (a[r] + 1) * (n - r) + k - (r - l + 1);
      if (ans > tmp) ans = tmp;
    }
  }
  cout << ans;
}
int main() {
  input();
  init();
  work();
  return 0;
}