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 = 2e5 + 7;
using ll = long long;
int a[N];
ll pre[N];
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
  ll ans = LLONG_MAX;
  for (int l = 1, r; l <= n; l = r) {
    r = l;
    while (r <= n && a[r] == a[l]) r++;
    ll cnt = r - l;
    if (cnt >= k) {
      ans = 0;
      break;
    }
    ll need = k - cnt;
    ll l_target = a[l] - 1;
    ll l_cnt = l - 1;
    ll l_base = l_cnt * l_target - pre[l - 1];
    ll r_target = a[l] + 1;
    ll r_cnt = n + 1 - r;
    ll r_base = pre[n] - pre[r - 1] - r_cnt * r_target;
    if (l_cnt >= need) {
      ans = min(ans, l_base + need);
    }
    if (r_cnt >= need) {
      ans = min(ans, r_base + need);
    }
    ans = min(ans, l_base + r_base + need);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}