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("O3")
#pragma GCC optimize("Ofast")
using namespace std;
long long int n, k, a[200005], dp[200005];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (long long int i = 1; i <= n; i++) dp[i] = dp[i - 1] + a[i];
  long long int i = 1, ans = -1;
  while (i <= n) {
    long long int j = i;
    while (j + 1 <= n && a[j + 1] == a[j]) {
      j++;
    }
    long long int cnt = j - i + 1;
    if (cnt >= k) {
      ans = 0;
      break;
    }
    long long int cur = 1e18;
    long long int s1 = i - 1;
    long long int s2 = n - j;
    long long int ans1 = 0, ans2 = 0;
    ans1 = s1 * (a[i] - 1) - dp[i - 1] + dp[0];
    ans2 = dp[n] - dp[j] - s2 * (a[i] + 1);
    if (s2 + cnt >= k) {
      cur = min(cur, ans2);
    }
    if (s1 + cnt >= k) {
      cur = min(cur, ans1);
    }
    cur = min(cur, ans1 + ans2);
    cur = cur + k - cnt;
    if (ans == -1) {
      ans = cur;
    } else
      ans = min(ans, cur);
    i = j + 1;
  }
  cout << ans << '\n';
}