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;
using ll = long long;
bool sortinrev(const pair<ll, ll> &a, const pair<ll, ll> &b) {
  if (a.first == b.first)
    return (a.second < b.second);
  else
    return (a.first < b.first);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    ll n, m;
    cin >> n >> m;
    vector<pair<ll, ll>> a;
    for (ll i = 0; i <= n - 1; ++i) {
      int tmp, cnt = 0;
      cin >> tmp;
      while (tmp > 0) {
        pair<ll, ll> lol;
        lol.first = tmp;
        lol.second = cnt;
        cnt++;
        tmp /= 2;
        a.push_back(lol);
      }
      pair<ll, ll> lol;
      lol.first = tmp;
      lol.second = cnt;
      a.push_back(lol);
    }
    sort(a.begin(), a.end(), sortinrev);
    ll now = a[0].first;
    ll numnow = 0;
    ll ans = 0;
    ll realans = INT_MAX;
    int i = 0;
    while (i < a.size()) {
      if (a[i].first == now) {
        numnow++;
        if (numnow <= m) ans += a[i].second;
        i++;
      } else {
        if (ans < realans && numnow >= m) realans = ans;
        ans = 0;
        now = a[i].first;
        numnow = 0;
      }
    }
    if (ans < realans && numnow >= m) realans = ans;
    cout << realans << "\n";
  }
}