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;
long long arr[200005];
map<long long, long long> mp;
map<long long, long long> mp1;
map<long long, long long> freq;
set<long long> s;
long long pre[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i < n + 1; i++)
    cin >> arr[i], freq[arr[i]]++, s.insert(arr[i]);
  sort(arr + 1, arr + n + 1);
  for (long long i = 1; i < n + 1; i++) pre[i] = pre[i - 1] + arr[i];
  long long sum = arr[1];
  long long c = 1;
  for (long long i = 2; i < n + 1; i++) {
    if (arr[i - 1] < arr[i]) mp[arr[i]] = sum, mp1[arr[i]] = c;
    sum += arr[i];
    c++;
  }
  long long ans = 1e16;
  for (auto i : s) {
    long long f = freq[i];
    if (f >= k) {
      cout << "0";
      return 0;
    }
    long long req = k - f;
    long long a = 0;
    long long x1 = (i - 1) * mp1[i] - mp[i];
    long long x2 =
        (sum - mp[i] - i * freq[i]) - (n - mp1[i] - freq[i]) * (i + 1);
    if (req <= mp1[i])
      a = x1 + req;
    else
      a = x1 + mp1[i] + x2 + (req - mp1[i]);
    ans = min(ans, a);
    a = 0;
    long long y = (n - mp1[i] - freq[i]);
    if (req <= y)
      a = x2 + req;
    else
      a = x1 + (req - y) + x2 + y;
    ans = min(ans, a);
  }
  cout << ans;
}