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 int N = (int)2e5 + 7;
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e18 + 7;
long long a[N], s[N];
int n, k;
long long sum(int l, int r) { return s[r] - s[l - 1]; }
long long proc(int l, int r) {
  int rem = k - (r - l + 1);
  if (rem <= 0) return 0;
  long long lef = l != 1 ? (a[l] - 1) * (l - 1) - sum(1, l - 1) : 0;
  long long rig = r != n ? sum(r + 1, n) - (a[r] + 1) * (n - r) : 0;
  long long afterL = k - r;
  long long afterR = k - (n - l + 1);
  long long ans1 = 1LL * min(rem, l - 1) + lef;
  if (afterL > 0) ans1 += rig + afterL;
  long long ans2 = 1LL * min(rem, n - r) + rig;
  if (afterR > 0) ans2 += lef + afterR;
  return min(ans1, ans2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + a[i];
  long long ans = INF;
  for (int l = 1; l <= n;) {
    int r = l;
    while (a[r + 1] == a[l] && r < n) r++;
    ans = min(ans, proc(l, r));
    l = r + 1;
  }
  cout << ans;
}