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;
template <class A, class B>
bool mini(A &a, const B b) {
  return a > b ? (a = b, 1) : 0;
}
template <class A, class B>
bool maxi(A &a, const B b) {
  return a < b ? (a = b, 1) : 0;
}
const long long maxN = 2e5 + 5;
const long long inf = 1e18;
long long k, n, a[maxN], p[maxN], s[maxN], res = inf, cp, cs, c;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = (1), _b = (n); i <= (_b); i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (long long i = (2), _b = (n); i <= (_b); i++) {
    p[i] = p[i - 1] + (a[i] - a[i - 1]) * (i - 1);
  }
  for (long long i = (n - 1), _b = (1); i >= (_b); i--) {
    s[i] = s[i + 1] + (a[i + 1] - a[i]) * (n - i);
  }
  for (long long j, i = 1; i <= n;) {
    for (j = i; i <= n && a[j] == a[i]; i++)
      ;
    if (i - j >= k) {
      res = 0;
      break;
    }
    c = k - (i - j);
    if (j > 1)
      cp = (a[j] - a[j - 1] - 1) * (j - 1) + p[j - 1];
    else
      cp = inf;
    if (i <= n)
      cs = (a[i] - a[j] - 1) * (n - i + 1) + s[i];
    else
      cs = inf;
    if (j - 1 >= c) mini(res, cp + c);
    if (n - i + 1 >= c) mini(res, cs + c);
    mini(res, cp + cs + c);
  }
  cout << res << '\n';
  return 0;
}