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;
using ld = long double;
using vi = vector<long long>;
using mi = map<long long, long long>;
using pi = pair<long long, long long>;
const long long N = 100005;
const long long MOD = 1e9 + 7;
const long long inf = 1e18 + 1;
void solve() {
  long long n, k, ans = inf;
  cin >> n >> k;
  long long a[n], pref[n + 1];
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  pref[0] = 0;
  for (long long i = 0; i < n; i++) pref[i + 1] = pref[i] + a[i];
  for (long long i = 0; i < n; i++) {
    long long x = a[i];
    long long lc = lower_bound(a, a + n, x) - a;
    long long mc = n - (upper_bound(a, a + n, x) - a);
    long long ls = pref[lc];
    long long ms = pref[n] - pref[n - mc];
    long long req = k - (n - lc - mc);
    if (req <= 0) {
      ans = 0;
      break;
    }
    if (mc >= req) {
      ans = min(ans, req + ms - mc * (x + 1));
    }
    if (lc >= req) {
      ans = min(ans, req + lc * (x - 1) - ls);
    }
    ans = min(ans, req + ms - mc * (x + 1) + lc * (x - 1) - ls);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}