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;
struct pair_hash {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2>& pair) const {
    return std::hash<T1>()(pair.first) ^ std::hash<T2>()(pair.second);
  }
};
const long double EPS = 1e-7;
const long long MOD = 1000000007;
const long long INF = 1e18;
const double pi = 2 * acos(0.0);
long long n, m, k, p, q, t, sum = 0, cnt = 0;
long long mx = INT64_MIN;
long long mn = INT64_MAX;
bool flag = 0;
vector<long long> v;
string s = "", s1, s2;
void bad_solve() {}
void solve() {
  vector<vector<long long> > nms(2e5 + 1, vector<long long>());
  cin >> n >> k;
  for (long long(i) = 0; (i) < (n); ++(i)) {
    cin >> p;
    cnt = 0;
    while (p) {
      nms[p].push_back(cnt);
      p >>= 1ll;
      ++cnt;
    }
    nms[p].push_back(cnt);
  }
  for (long long(i) = 0; (i) < (2e5 + 1); ++(i)) {
    sort(nms[i].begin(), nms[i].end());
    if ((long long)nms[i].size() >= k) {
      sum = 0;
      for (long long(j) = 0; (j) < (k); ++(j)) {
        sum += nms[i][j];
      }
      mn = min(sum, mn);
    }
  }
  cout << mn;
}
int main() {
  cout.precision(20);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}