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
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
inline int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
inline int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
inline int mul(int a, int b) { return 1ll * a * b % P; }
inline int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 404040;
int n, m;
int sum[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (n); i++) {
    int t;
    scanf("%d", &t);
    sum[t]++;
  }
  for (int i = (1); i < (N); i++) sum[i] = sum[i - 1] + sum[i];
  int ans = 1e9;
  for (int i = (1); i < (N / 2); i++) {
    int s = 0, op = 0, res = 0;
    for (int l = i, r = i; s < m && r < N; l = l * 2, r = r * 2 + 1, op++) {
      int t = min(m - s, sum[r] - sum[l - 1]);
      s += t;
      res += t * op;
    }
    if (s == m) ans = min(ans, res);
  }
  if (ans == 1e9) {
    ans = 0;
    for (int i = (1); i < (N / 2); i++) {
      int t = 0, sb = i;
      while (sb) {
        sb >>= 1;
        t++;
      }
      ans += t * (sum[i] - sum[i - 1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}
