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;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long a[n + 1];
  map<long long, long long> freq;
  map<long long, long long> first, last;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    freq[a[i]]++;
  }
  sort(a + 1, a + n + 1);
  long long pre[n + 1];
  pre[0] = 0;
  for (long long i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + a[i];
    if (!first[a[i]]) {
      first[a[i]] = i;
    }
    last[a[i]] = i;
  }
  map<long long, long long>::iterator it;
  long long res = LONG_LONG_MAX;
  for (it = freq.begin(); it != freq.end(); it++) {
    if (k <= (*it).second) {
      res = 0;
      break;
    }
    long long small = first[(*it).first] - 1;
    long long big = n - last[(*it).first];
    long long ctr = k - (*it).second;
    long long sum1 = pre[small] - pre[0];
    long long sum2 = pre[n] - pre[last[(*it).first]];
    sum1 = small * ((*it).first - 1) - sum1;
    sum2 -= big * ((*it).first + 1);
    if ((ctr <= small) && (ctr <= big)) {
      long long sum = min(sum1, sum2);
      sum += ctr;
      res = min(res, sum);
    } else if ((ctr <= small)) {
      long long sum = sum1;
      sum += ctr;
      res = min(res, sum);
    } else if (ctr <= big) {
      long long sum = sum2;
      sum += ctr;
      res = min(res, sum);
    } else {
      long long sum = sum1 + sum2;
      sum += ctr;
      res = min(res, sum);
    }
  }
  if (res < 0) {
    res = 0;
  }
  cout << res << "\n";
}