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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k, n, m, p, q, t, ans = LLONG_MAX, sum = 0, ct = 0, flag = 0;
  cin >> n >> k;
  std::vector<long long> v(n);
  multiset<long long> st;
  std::vector<long long> v1, v2;
  for (i = 0; i < n; i++) {
    cin >> v[i];
    st.insert(v[i]);
  }
  sort(v.begin(), v.end());
  long long cost = 0, pct, low, high;
  auto it1 = st.end();
  long long num;
  auto it2 = st.end();
  for (j = 0; j < 200000; j++) {
    cost = 0;
    q = k;
    for (i = 0; i < 20; i++) {
      pct = (long long)pow(2, i);
      low = pct * j;
      high = low + pct - 1;
      low--;
      if (j == 0 && i > 0) {
        low = (long long)pow(2, i - 1) - 1;
        high = (long long)pow(2, i) - 1;
      }
      it2 = st.upper_bound(high);
      it1 = st.upper_bound(low);
      int dist = distance(it1, it2);
      if (dist >= q) {
        cost += q * i;
        q = 0;
        break;
      } else {
        cost += dist * i;
        q -= dist;
      }
      if (q <= 0) break;
    }
    if (cost <= ans && q == 0) {
      ans = cost;
      num = j;
    }
  }
  cout << ans;
  return 0;
}