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 long long N = 200005;
long long n, k, a[N], pre[N], suf[N], ans = 1e18;
map<long long, long long> mp;
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + a[i];
  }
  for (long long i = n; i >= 1; --i) {
    suf[i] = suf[i + 1] + a[i];
  }
  for (long long i = 1; i <= n; i++) mp[a[i]]++;
  for (auto i : mp)
    if (i.second >= k) {
      cout << 0 << endl;
      return 0;
    }
  for (long long i = 1; i <= n; i++) {
    long long x = a[i];
    long long e =
        upper_bound(a + 1, a + n + 1, x) - lower_bound(a + 1, a + n + 1, x);
    long long g = n - (upper_bound(a + 1, a + n + 1, x) - a) + 1;
    long long l = lower_bound(a + 1, a + n + 1, x) - a - 1;
    long long sg = suf[n - g + 1];
    long long sl = pre[l];
    long long tmp = sg - g * x + l * x - sl - (n - k);
    ans = min(ans, tmp);
    if (n - l >= k) {
      long long tx = sg - g * x - (n - l - k);
      ans = min(ans, tx);
    }
    if (n - g >= k) {
      long long tx = l * x - sl - (n - g - k);
      ans = min(ans, tx);
    }
  }
  cout << max(0ll, ans);
}