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 int;
ll d, x, y;
ll power(ll x, ll n, ll mod = 1e9 + 7) {
  ll result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result % mod * x % mod) % mod;
    x = (x % mod * x % mod) % mod;
    n = n / 2;
  }
  return result;
}
void gcd(ll a, ll b) {
  if (b == 0)
    d = a, x = 1, y = 0;
  else {
    gcd(b, a % b);
    ll temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
ll modInverse(ll a, ll m) {
  gcd(a, m);
  return (x % m + m) % m;
}
vector<ll> factors(ll n) {
  vector<ll> res;
  for (ll i = 2; i <= sqrt(n); i++) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  return res;
}
int main() {
  ll n, k;
  cin >> n >> k;
  vector<ll> a(n);
  for (ll &it : a) cin >> it;
  sort(a.begin(), a.end());
  ll mx = *max_element(a.begin(), a.end());
  ll ans = INT_MAX;
  for (ll i = 0; i <= mx; i++) {
    ll c = 0;
    ll x = 0;
    for (ll it : a) {
      ll temp = it;
      ll t = 0;
      while (temp > i) {
        temp = temp / 2;
        t++;
      }
      if (temp == i) {
        c++;
        x += t;
      }
      if (c == k) {
        break;
      }
    }
    if (c == k) ans = min(ans, x);
  }
  cout << ans << endl;
}