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 int inf = 0x3f3f3f3f;
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts>
void err(const T& arg, const Ts&... args) {
  cout << arg << " ";
  err(args...);
}
const int N = 200000 + 5;
int n, k, a[N];
vector<int> v;
long long cnt[N], sum[N];
long long pre[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    v.push_back(a[i]);
  }
  v.push_back(0);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; i++) {
    int id = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
    cnt[id]++;
  }
  n = v.size() - 1;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + cnt[i];
    pre[i] = pre[i - 1] + cnt[i] * v[i];
  }
  long long res = pre[n];
  for (int i = 1; i <= n; i++) {
    long long need = max(0ll, k - cnt[i]);
    {
      long long tmp = 0;
      int lnum = min(need, sum[i - 1]);
      if (lnum) {
        tmp = sum[i - 1] * (v[i] - 1) - pre[i - 1];
        tmp += lnum;
      }
      int rnum = min(need - lnum, sum[n] - sum[i]);
      if (rnum) {
        tmp += (pre[n] - pre[i]) - (sum[n] - sum[i]) * (v[i] + 1);
        tmp += rnum;
      }
      res = min(res, tmp);
    }
    {
      long long tmp = 0;
      int rnum = min(need, sum[n] - sum[i]);
      if (rnum) {
        tmp = (pre[n] - pre[i]) - (sum[n] - sum[i]) * (v[i] + 1);
        tmp += rnum;
      }
      int lnum = min(need - rnum, sum[i - 1]);
      if (lnum) {
        tmp += sum[i - 1] * (v[i] - 1) - pre[i - 1];
        tmp += lnum;
      }
      res = min(res, tmp);
    }
  }
  cout << res << endl;
  return 0;
}