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>
const double eps = 1e-8, pi = acos(-1.0);
const int maxS = 1e6 + 13;
const int maxT = 1e3 + 13;
const int MOD = 835672545;
const int base = 19260817;
const int layer = 20;
using namespace std;
inline void read(char &x) { scanf("%c", &x); }
inline void read(long long &x) { scanf("%lld", &x); }
inline void read(long long &x, long long &y) {
  read(x);
  read(y);
}
inline void read(long long &x, long long &y, long long &z) {
  read(x, y);
  read(z);
}
inline void read(int &x) { scanf("%d", &x); }
inline void read(int &x, int &y) {
  read(x);
  read(y);
}
inline void read(int &x, int &y, int &z) {
  read(x, y);
  read(z);
}
int n, m, x, y, z, k, T, kase = 0;
int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};
int a[maxS], res[maxS], cnt[maxS];
int main() {
  read(n, m);
  for (int i = 0; i < n; i++) read(a[i]);
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int x = a[i], k = 0;
    while (x >= 0) {
      if (cnt[x] == m) break;
      cnt[x]++;
      res[x] += k;
      k++;
      if (x)
        x /= 2;
      else
        x = -1;
    }
  }
  int ans = 1e9;
  for (int i = 0; i <= 2e5; i++)
    if (cnt[i] == m) ans = min(ans, res[i]);
  printf("%d\n", ans);
  return 0;
}