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;
long long n, k;
void search(int cur) {
  int tot, c[10];
  int ok;
  if (cur == n)
    tot++;
  else
    for (int i = 0; i < n; i++) {
      ok = 1;
      c[cur] = i;
    }
  for (int j = 0; j < cur; j++) {
    if (c[cur] == c[j] || cur - c[cur] == j - c[j] ||
        cur + c[cur] == j + c[j]) {
      ok = 0;
      break;
    }
  }
  if (ok) search(cur + 1);
}
const long long MAXN = 55;
long long num[MAXN];
void xswl() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld", &num[i]);
  }
  num[0] = 1;
  sort(num + 1, num + 1 + n);
  long long res = 999999999999999;
  for (long long i = 2e5; i >= 0; --i) {
    long long cnt = 0;
    long long nowr = 0;
    int t = lower_bound(num + 1, num + 1 + n, i) - num;
    for (int j = t; j <= n; ++j) {
      long long now = 0;
      long long u = num[j];
      while (u > i) {
        ++now;
        u >>= 1;
      }
      if (u == i) {
        ++cnt;
        nowr += now;
      }
      if (cnt == k) break;
    }
    if (cnt == k) {
      if (res > nowr) {
        res = nowr;
      }
    }
  }
  printf("%lld", res);
}
int main() { xswl(); }