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 N = 2e5 + 10;
int n, k, A[N], maxn, Ans;
vector<int> Cost[N];
inline int read() {
  int rt = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    rt = (rt << 3) + (rt << 1) + (ch & 15);
    ch = getchar();
  }
  return rt;
}
inline void add(int x) {
  int Cnt = 0;
  while (x) Cost[x].push_back(Cnt++), x >>= 1;
}
int main() {
  n = read(), k = read();
  for (register int i = 1; i <= n; ++i)
    A[i] = read(), add(A[i]), maxn = max(maxn, A[i]);
  Ans = 1e9;
  int Cnt, siz;
  for (register int i = 1; i <= maxn; ++i) {
    siz = Cost[i].size();
    if (siz < k) continue;
    sort(Cost[i].begin(), Cost[i].end());
    Cnt = 0;
    for (register int j = 0; j <= k - 1; ++j) Cnt += Cost[i][j];
    Ans = min(Ans, Cnt);
  }
  return not printf("%d", Ans);
}
