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 N = 200001;
const int K = 441;
long long n, k;
long long a[N];
long long cnt;
inline long long solve_up() {
  long long x = a[k];
  long long cnt = 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] == x) ++cnt;
  long long left = k - cnt;
  long long at_least = x - 1;
  long long ans = 0;
  for (long long i = 1; i < k; ++i) {
    if (a[i] <= at_least) ans += at_least - a[i];
  }
  ans += left;
  return ans;
}
inline long long solve_down() {
  long long x = a[n - k + 1];
  long long cnt = 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] == x) ++cnt;
  long long left = k - cnt;
  long long at_least = x + 1;
  long long ans = 0;
  for (long long i = n - k + 1; i <= n; ++i) {
    if (a[i] >= at_least) ans += a[i] - at_least;
  }
  ans += left;
  return ans;
}
long long r[N];
map<long long, long long> c, res;
inline long long solve_up_and_down() {
  long long s = 0;
  long long ans = 1e18;
  for (long long i = 1; i <= n; ++i) {
    if (a[i] == a[i - 1]) {
      s += a[i];
      continue;
    }
    long long at_least = a[i] - 1;
    long long cur_ans = at_least * (i - 1) - s;
    res[a[i]] = cur_ans;
    s += a[i];
  }
  s = 0;
  for (long long i = n; i >= 1; --i) {
    if (a[i] == a[i + 1]) {
      s += a[i];
      continue;
    }
    long long at_least = a[i] + 1;
    long long cur_ans = s - at_least * (n - i);
    res[a[i]] += cur_ans;
    s += a[i];
  }
  for (auto &x : res) {
    long long key = x.first, value = x.second;
    long long cnt = c[key];
    long long left = k - cnt;
    ans = min(ans, value + left);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i], c[a[i]]++;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (a[i] == a[i - 1])
      ++cnt;
    else
      cnt = 1;
    if (cnt == k) return cout << 0, 0;
  }
  long long ans = solve_up();
  ans = min(ans, solve_down());
  ans = min(ans, solve_up_and_down());
  cout << ans;
  return 0;
}