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 N = 1e6 + 100;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long llINF = 0x3f3f3f3f3f3f3f3f;
inline bool read(long long &num) {
  char in;
  bool IsN = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else
    num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') {
    num *= 10, num += in - '0';
  }
  if (IsN) num = -num;
  return true;
}
long long a[N], x, n, k;
set<long long> st;
long long num[N], ti[N];
int main() {
  read(n);
  read(k);
  for (int i = (1); i <= (n); i++) read(a[i]);
  sort(a + 1, a + n + 1);
  long long ans = 9999999999999;
  for (int i = (1); i <= (n); i++) {
    long long x = a[i];
    int cnt = 0;
    while (x) {
      num[x]++;
      ti[x] += cnt;
      if (num[x] >= k) ans = min(ans, ti[x]);
      cnt++;
      x /= 2;
    }
  }
  cout << ans << endl;
  return 0;
}