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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll getForwardAns(vector<int> arr, int k) {
  map<int, int> eqCountIfTakenHere;
  map<int, ll> stepsNeeded;
  map<int, int> outsiders;
  int n = arr.size();
  vector<int> outsiderPerElem(n);
  vector<ll> forwardCC(n);
  for (ll i = 1; i < n; ++i) {
    forwardCC[i] = forwardCC[i - 1] + i * abs(arr[i] - arr[i - 1]);
  }
  for (int i = 0; i < n; ++i) {
    if (i > 0 && arr[i] != arr[i - 1])
      outsiderPerElem[i] = i;
    else if (i > 0)
      outsiderPerElem[i] = outsiderPerElem[i - 1];
  }
  for (int i = 0; i < n; ++i) {
    eqCountIfTakenHere[arr[i]] = i + 1;
    stepsNeeded[arr[i]] = forwardCC[i];
    outsiders[arr[i]] = outsiderPerElem[i];
  }
  ll ans = 1e18;
  for (int x : arr) {
    if (eqCountIfTakenHere[x] < k) continue;
    int core = eqCountIfTakenHere[x] - outsiders[x];
    if (core >= k) {
      ans = 0;
    } else {
      int needed = k - core;
      ans = min(ans, stepsNeeded[x] - outsiders[x] + needed);
    }
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  sort(begin(arr), end(arr));
  ll ans = 0;
  int medOutsiders = 0;
  for (int i = 0; i < n; ++i) {
    ans += abs(arr[i] - arr[n / 2]);
    if (arr[i] != arr[n / 2]) ++medOutsiders;
  }
  if (n - medOutsiders < k) {
    int needed = k - (n - medOutsiders);
    ans += (needed - medOutsiders);
  } else {
    ans = 0;
  }
  ans = min(ans, getForwardAns(arr, k));
  reverse(begin(arr), end(arr));
  ans = min(ans, getForwardAns(arr, k));
  printf("%lld\n", ans);
  return 0;
}
