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 <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
T add(T &a, T b) {
  return (a += b) %= 1000000007;
}
template <typename T>
T sub(T &a, T b) {
  (a -= b);
  if (a < 0) a += 1000000007;
}
const int N = 2e5 + 5;
long long n, k;
long long a[N], b[N], ans;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (long long i = 0; i < n; i++) b[i + 1] = b[i] + a[i];
  ans = 9e18;
  for (long long i = 0, j = 0; i < n; i = j) {
    for (; j < n and a[i] == a[j]; j++)
      ;
    long long c = j - i, first = a[i], cl = i * (first - 1) - b[i],
              cr = b[n] - b[j] - (n - j) * (first + 1);
    if (c >= k) ans = 0;
    if (c + i >= k) ans = min(ans, cl + k - c);
    if (c + n - j >= k) ans = min(ans, cr + k - c);
    ans = min(ans, (cl + cr + k - c));
  }
  cout << max(ans, 0ll);
  return 0;
}