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 int oo = 1e9 + 7;
const int modulo = 1e9 + 7;
template <typename T>
inline T sum(const T& a, const T& b) {
  return T((a % modulo + b % modulo) % modulo);
}
template <typename T>
inline T mul(const T& a, const T& b) {
  return T((1LL * (a % (1LL * modulo)) * (b % (1LL * modulo))) %
           (1LL * modulo));
}
string to_string(string a) { return "\"" + a + "\""; }
const int N = 2e5 + 5;
int n, k, cnt[N];
vector<pair<int, int> > g[N];
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  memset(cnt, 0, sizeof cnt);
  for (int i = 1; i <= n; ++i) {
    int a;
    cin >> a;
    cnt[a]++;
  }
  int ans = oo;
  for (int i = N - 1; i >= 0; --i) {
    if (cnt[i]) {
      int tmp = i, r = 0;
      while (tmp) {
        g[tmp].push_back(pair<int, int>(cnt[i], r));
        tmp /= 2;
        r++;
      }
      g[0].push_back(pair<int, int>(cnt[i], r));
    }
  }
  for (int i = 0; i < N; ++i)
    if (int(g[i].size()) != 0) {
      int tmp = 0, sz = 0;
      sort(g[i].begin(), g[i].end(), cmp);
      for (pair<int, int> v : g[i]) {
        int cmin = min(k - sz, v.first);
        sz += cmin;
        tmp += cmin * v.second;
      }
      if (sz == k) {
        ans = min(ans, tmp);
      }
    }
  if (ans == oo)
    cout << n;
  else
    cout << ans;
  return 0;
}