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;
int n, k;
int a[202202];
pair<int, int> p[202202];
int lp;
long long int calc1(int fr, int step) {
  long long int res = 0LL;
  int cnt = p[fr].second;
  for (int i = fr; i < lp && i >= 0; i += step) {
    int cur = i, nxt = i + step;
    res += 1LL * cnt * (labs(p[cur].first - p[nxt].first) - 1);
    if (cnt + p[nxt].second < k) {
      res += cnt;
      cnt += p[nxt].second;
    } else {
      res += k - p[nxt].second;
      break;
    }
  }
  return res;
}
long long int calc2(int l, int r) {
  long long int res = 0LL;
  int lef = 0, rig = 0;
  for (int i = 0; i < lp; i++) {
    if (p[i].first <= l) {
      res += 1LL * (l - p[i].first) * p[i].second;
      lef += p[i].second;
    } else {
      res += 1LL * (p[i].first - r) * p[i].second;
      rig += p[i].second;
    }
  }
  return res + k - max(lef, rig);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  lp = 0;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (i > 0 && a[i] != a[i - 1]) {
      p[lp++] = make_pair(a[i - 1], cnt);
      cnt = 1;
    } else
      cnt++;
  }
  p[lp++] = make_pair(a[n - 1], cnt);
  bool chk = false;
  for (int i = 0; i < lp && !chk; i++) {
    if (p[i].second >= k) chk = true;
  }
  if (chk) {
    puts("0");
    return 0;
  }
  long long int ans = calc1(0, 1);
  ans = min(ans, calc1(lp - 1, -1));
  if (n % 2 == 0)
    ans = min(ans, calc2(a[n / 2] - 1, a[n / 2]));
  else {
    ans = min(ans, calc2(a[n / 2] - 1, a[n / 2]));
    ans = min(ans, calc2(a[n / 2], a[n / 2] + 1));
  }
  printf("%I64d\n", ans);
  return 0;
}