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;
double const EPS = 1e-12, PI = acos(-1);
const int N = 2e5 + 9, M = 1e7 + 9, OO = 1e9 + 7, MOD = 1e9 + 7,
          MOD2 = 998244353;
const long long INF = 1e18;
map<int, int> mp;
pair<int, int> arr[N];
long long pref[N], suf[N];
int prefCnt[N], sufCnt[N];
int main() {
  cout << fixed << setprecision(12), ios::sync_with_stdio(false),
      cin.tie(nullptr), cout.tie(nullptr);
  int n, k, a;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    mp[a]++;
  }
  int cur = 1;
  for (auto it : mp) arr[cur++] = it;
  long long cnt = 0, cntNum = 0;
  for (int i = 1; i <= cur; ++i) {
    cntNum += arr[i].second;
    cnt += 1ll * arr[i].first * arr[i].second;
    pref[i] = cnt;
    prefCnt[i] = cntNum;
  }
  cnt = 0, cntNum = 0;
  for (int i = cur; i > 0; --i) {
    cntNum += arr[i].second;
    cnt += 1ll * arr[i].first * arr[i].second;
    suf[i] = cnt;
    sufCnt[i] = cntNum;
  }
  long long ans = INF;
  for (int i = 1; i < cur; ++i) {
    int req = k - arr[i].second;
    if (req <= 0) {
      ans = 0;
      break;
    }
    int cntBefore = min(req, prefCnt[i - 1]);
    long long c1 = 1ll * prefCnt[i - 1] * (arr[i].first - 1) - pref[i - 1];
    c1 += cntBefore;
    int cntNxt = req - cntBefore;
    assert(cntNxt >= 0);
    if (cntNxt) {
      c1 += suf[i + 1] - 1ll * sufCnt[i + 1] * (arr[i].first + 1);
      c1 += cntNxt;
    }
    ans = min(ans, c1);
    cntNxt = min(req, sufCnt[i + 1]);
    c1 = suf[i + 1] - 1ll * sufCnt[i + 1] * (arr[i].first + 1);
    c1 += cntNxt;
    cntBefore = req - cntNxt;
    assert(cntBefore >= 0);
    if (cntBefore) {
      c1 += 1ll * prefCnt[i - 1] * (arr[i].first - 1) - pref[i - 1];
      c1 += cntBefore;
    }
    ans = min(ans, c1);
  }
  cout << ans;
  return 0;
}