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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long N = 2e5 + 7;
long long n, k;
long long p[N], s[N];
long long get_p(long long x) {
  if (x < 0) {
    return 0;
  }
  return p[x];
}
long long get_s(long long x) {
  if (x >= n) {
    return 0;
  }
  return s[x];
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  for (long long i = 0; i < n; i++) {
    p[i] = a[i] + (i ? p[i - 1] : 0);
  }
  for (long long i = n - 1; i >= 0; i--) {
    s[i] = a[i] + (i != n - 1 ? s[i + 1] : 0);
  }
  long long lst = -1;
  long long ans = 1e18 + 7;
  for (long long i = 0; i < n; i++) {
    if (i == 0 || a[i] != a[i - 1]) {
      lst = i;
    }
    if (i + 1 < n && a[i + 1] == a[i]) {
      continue;
    }
    long long cnt = i - lst + 1;
    if (cnt >= k) {
      cout << 0 << '\n';
      return 0;
    }
    long long cnt_l = lst;
    long long cnt_r = n - i - 1;
    long long need = k - cnt;
    if (cnt_l < need) {
      long long sum = cnt_l * a[i];
      long long have = need - cnt_l;
      sum = sum - get_p(lst - 1);
      long long kek = have * a[i];
      kek += (cnt_r - have) * (a[i] + 1);
      kek = get_s(i + 1) - kek;
      ans = min(ans, kek + sum);
    } else {
      long long sum = need * a[i];
      sum += (cnt_l - need) * (a[i] - 1);
      sum = sum - get_p(lst - 1);
      ans = min(ans, sum);
    }
    if (cnt_r < need) {
      long long sum = cnt_r * a[i];
      long long have = need - cnt_r;
      sum = get_s(i + 1) - sum;
      long long kek = have * a[i];
      kek += (cnt_l - have) * (a[i] - 1);
      kek = kek - get_p(lst - 1);
      ans = min(ans, kek + sum);
    } else {
      long long sum = need * a[i];
      sum += (cnt_r - need) * (a[i] + 1);
      sum = get_s(i + 1) - sum;
      ans = min(ans, sum);
    }
  }
  cout << ans << '\n';
  return 0;
}