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 long long N = 400010;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long n, K, a[N], qz[N], hz[N], Ans = 1e18;
signed main() {
  n = read(), K = read();
  for (register long long i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + 1 + n);
  long long now = 0;
  for (register long long i = 1; i <= n; i++) {
    if (i == 1 || a[i] == a[i - 1])
      now++;
    else {
      if (now >= K) return puts("0") & 0;
      now = 1;
    }
  }
  for (register long long i = 1; i <= n; i++) qz[i] = qz[i - 1] + a[i];
  for (register long long i = n; i; i--) hz[i] = hz[i + 1] + a[i];
  for (register long long i = 1; i <= n; i++) {
    if (i >= K) Ans = min(Ans, a[i] * i - qz[i] - (i - K));
    if (n - i + 1 >= K)
      Ans = min(Ans, hz[i] - a[i] * (n - i + 1) - (n - i + 1 - K));
    Ans = min(Ans, hz[i] - a[i] * (n - i + 1) - qz[i] + a[i] * i - (n - K));
  }
  printf("%lld\n", max(Ans, 0ll));
  return 0;
}