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
#include <bits/stdc++.h>
using namespace std;
const long long N = 200007;
vector<long long> v[N + 5];
long long cnt[N + 9];
long long a[N + 9];
set<long long> st;
int32_t main() {
  ios::sync_with_stdio(0);
  long long n, k, x;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> x;
    ++cnt[x];
    if (cnt[x] >= k) {
      st.insert(x);
    }
    long long r = 0;
    v[x].push_back(r);
    while (x != 0) {
      x /= 2;
      r++;
      ++cnt[x];
      if (cnt[x] >= k) {
        st.insert(x);
      }
      v[x].push_back(r);
    }
  }
  long long ans = 999999999999999999;
  for (auto it : st) {
    sort(v[it].begin(), v[it].end());
    long long r = 0;
    for (long long i = 0; i < k; i++) {
      r += v[it][i];
    }
    ans = min(ans, r);
  }
  cout << ans;
  return 0;
}
