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;
using ll = long long;
struct Tree {
  static const ll LOW = 0;
  ll f(ll a, ll b) { return a + b; }
  vector<ll> s;
  int n;
  Tree(int n = 0, ll def = 0) : s(2 * n, def), n(n) {}
  void update(int pos, ll val) {
    for (s[pos += n] = val; pos > 1; pos /= 2)
      s[pos / 2] = f(s[pos & (~1)], s[pos | 1]);
  }
  ll query(int b, int e) {
    if (b >= e) return 0;
    ll ra = LOW, rb = LOW;
    for (b += n, e += n; b < e; b /= 2, e /= 2) {
      if (b % 2) ra = f(ra, s[b++]);
      if (e % 2) rb = f(s[--e], rb);
    }
    return f(ra, rb);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll n, k;
  cin >> n >> k;
  Tree rage(n, 1);
  ll a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) rage.update(i, a[i]);
  ll out = ((ll)1) << 62;
  ll cnt = 1;
  ll begin = 0;
  for (int i = 0; i < n; i++) {
    if (i >= 1 && a[i - 1] != a[i]) {
      cnt = 1;
      begin = i;
    }
    if (i < n - 1 && a[i] == a[i + 1]) {
      cnt++;
      continue;
    }
    ll local = ((ll)1) << 62;
    ll needed = k - cnt;
    ll target = a[i];
    if (needed <= 0) {
      cout << 0 << endl;
      return 0;
    }
    ll leftside = rage.query(0, begin);
    leftside = abs(begin * (target - 1) - leftside);
    ll rightside = rage.query(i + 1, n);
    rightside = abs(rightside - (n - (i + 1)) * (target + 1));
    local = min(local, leftside + rightside + needed);
    if (begin >= needed) {
      local = min(local, leftside + needed);
    }
    if (i < (n - needed)) {
      local = min(local, rightside + needed);
    }
    out = min(out, local);
  }
  cout << out << endl;
}