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;
template <class T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
}
const int N = 2e5 + 10;
long long a[N];
long long num[N];
long long d[N];
long long len[N];
vector<int> v[N];
bool cmp(long long a, long long b) { return a > b; }
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  {
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      num[a[i]]++;
    }
    for (int i = 0; i < 2e5; i++)
      if (num[i] >= k) return cout << 0 << endl, 0;
    int ans = 2e9;
    for (int i = 1; i <= n; i++) {
      d[i] = a[i];
      int cnt = 1;
      while (d[i]) {
        d[i] /= 2;
        v[d[i]].push_back(cnt);
        cnt++;
      }
      v[0].push_back(cnt);
      len[i] = cnt;
    }
    for (int i = 0; i <= 2e5; i++) sort(v[i].begin(), v[i].end());
    for (int i = 0; i < 2e5; i++) {
      int numm = num[i];
      int cha = k - numm;
      int Max = 0;
      int j = 0;
      while (j < v[i].size() && cha > 0) {
        Max += v[i][j++];
        cha--;
      }
      if (cha <= 0) ans = min(ans, Max);
    }
    cout << ans << endl;
  }
  return 0;
}