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>
#pragma warning(disable : 4996)
using namespace std;
const unsigned long long nmax = 200002;
unsigned long long n, k, a[nmax], s[nmax], t[nmax], c[nmax], r = UINT64_MAX, A,
                                                             B;
int main() {
  scanf("%llu%llu", &n, &k);
  for (unsigned long long i = 1; i <= n; ++i) scanf("%llu", a + i);
  sort(a + 1, a + n + 1);
  for (unsigned long long i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i];
  for (unsigned long long i = n; i >= 1; --i) t[i] = t[i + 1] + a[i];
  for (unsigned long long i = 1; i <= n; ++i) {
    if (a[i] == a[i - 1])
      c[i] = c[i - 1] + 1;
    else
      c[i] = 1;
    if (c[i] >= k) {
      puts("0");
      return 0;
    }
  }
  for (unsigned long long i = 1; i <= n; ++i) {
    if (i >= k) {
      A = i * a[i] - s[i] - (i - k);
      r = min(r, A);
    }
    if (n - i + 1 >= k) {
      B = t[i] - (n - i + 1) * a[i] - (n - i + 1 - k);
      r = min(r, B);
    }
    r = min(r, i * a[i] - s[i] + t[i] - (n - i + 1) * a[i] - (n - k));
  }
  printf("%llu\n", r);
  return 0;
}