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;
int a[1919810], m[1919810], s[1919810], n, k, minn = 1234567890, maxx;
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    maxx = max(maxx, a[i]);
    m[a[i]]++;
    if (m[a[i]] == k) exit(0 & puts("0"));
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) {
    int step = 0;
    while (a[i] != 0) {
      a[i] >>= 1, ++step;
      if (m[a[i]] == k) continue;
      ++m[a[i]], s[a[i]] += step;
    }
    if (a[i] == 0) ++m[0], s[0] += step;
  }
  int ans = 999999999;
  for (int i = 1; i <= maxx; i++)
    if (m[i] >= k) ans = min(ans, s[i]);
  printf("%d", ans);
  return 0;
}