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;
void fakemain() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  long long sum = 0;
  map<long long, long long> m;
  set<long long> s;
  long long ans = LLONG_MAX;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sum += x;
    m[x]++;
    if (m[x] >= k) ans = 0;
    s.insert(x);
  }
  if (ans == 0) {
    cout << 0 << "\n";
    return;
  }
  long long cnt = 0;
  long long pre = 0;
  for (auto it : m) {
    long long f = it.second;
    long long num = it.first;
    if (f >= k) {
      ans = 0;
      break;
    }
    if (cnt + f >= k) {
      long long req = cnt * (num - 1) - pre;
      req += (k - f);
      ans = min(ans, req);
    }
    if (n - cnt >= k) {
      long long req = (sum - pre - f * num) - (n - cnt - f) * (num + 1);
      req += (k - f);
      ans = min(ans, req);
    }
    long long req = cnt * (num - 1) - pre + (sum - pre - f * num) -
                    (n - cnt - f) * (num + 1);
    req += (k - f);
    ans = min(ans, req);
    cnt += f;
    pre += f * num;
  }
  cout << ans;
}
int main() { fakemain(); }