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;
const long long MOD = 1000000007;
const long long MAX = 2e6 + 7;
const long long INF = 1e15;
const long long dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const long long dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
long long i, j, k, t, n, q, l, r, mid;
vector<pair<long long, long long> > v;
long long precnt[MAX], presum[MAX], sufcnt[MAX], sufsum[MAX];
long long x, res, need, needr, needl, mini;
map<long long, long long> mp;
void solve() {
  mini = INF;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    mp[x]++;
  }
  for (auto x : mp) {
    v.push_back({x.first, x.second});
  }
  n = v.size();
  for (long long i = 0; i < n; i++) {
    if (i) {
      precnt[i] += precnt[i - 1];
      presum[i] += presum[i - 1];
    }
    precnt[i] += v[i].second;
    presum[i] += v[i].second * v[i].first;
  }
  for (long long i = n - 1; i >= 0; i--) {
    if (i != n - 1) {
      sufcnt[i] += sufcnt[i + 1];
      sufsum[i] += sufsum[i + 1];
    }
    sufcnt[i] += v[i].second;
    sufsum[i] += v[i].first * v[i].second;
  }
  for (long long i = 0; i < n; i++) {
    need = max(0ll, k - v[i].second);
    res = 0;
    needl = needr = 0;
    if (i) needl = min(need, precnt[i - 1]);
    needr = max(0ll, need - needl);
    if (i && needl) {
      res += (v[i].first - 1) * precnt[i - 1] - presum[i - 1];
      res += needl;
    }
    if (i + 1 < v.size() && needr) {
      res += sufsum[i + 1] - (v[i].first + 1) * sufcnt[i + 1];
      res += needr;
    }
    mini = min(mini, res);
    res = needl = needr = 0;
    if (i + 1 < v.size()) needr = min(need, sufcnt[i + 1]);
    needl = max(0ll, need - needr);
    if (i && needl) {
      res += (v[i].first - 1) * precnt[i - 1] - presum[i - 1];
      res += needl;
    }
    if (i + 1 < v.size() && needr) {
      res += sufsum[i + 1] - (v[i].first + 1) * sufcnt[i + 1];
      res += needr;
    }
    mini = min(mini, res);
  }
  cout << mini << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}