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;
const long long N = 2e6 + 7;
const long long INF = 1e18;
long long n, k, a[N], pre[N], ans = INF;
map<long long, long long> mp;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + a[i];
    mp[a[i]]++;
    if (mp[a[i]] >= 2) {
      a[i] = 0;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (a[i] == 0) {
      continue;
    }
    long long need = k - mp[a[i]];
    if (need <= 0) {
      cout << 0;
      return 0;
    }
    long long q_up = n - i - mp[a[i]] + 1;
    long long q_down = i - 1;
    long long cost_up =
        (pre[n] - pre[i]) - (mp[a[i]] - 1) * a[i] - q_up * (a[i] + 1);
    long long cost_down = (a[i] - 1) * (i - 1) - pre[i - 1];
    if (q_up >= need) {
      ans = min(ans, need + cost_up);
    } else {
      ans = min(ans, need + cost_up + cost_down);
    }
    if (q_down >= need) {
      ans = min(ans, need + cost_down);
    } else {
      ans = min(ans, need + cost_down + cost_up);
    }
  }
  cout << ans;
}