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 MAXN = 2e5 + 10;
const long long INF = 1e18;
const long long mod = 1e9 + 7;
long long n, k, ans = INF;
long long a[MAXN], sum[MAXN];
long long cnt[MAXN], l[MAXN], r[MAXN];
bool flag = false;
vector<long long> v;
inline long long Getid(long long x) {
  return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
signed main() {
  cin >> n >> k;
  for (register long long i = 1; i <= n; ++i) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  sort(a + 1, a + n + 1);
  v.erase(unique(v.begin(), v.end()), v.end());
  for (register long long i = 1; i <= n; ++i) {
    long long id = Getid(a[i]);
    if (a[i - 1] != a[i]) l[id] = i;
    r[id] = i;
    sum[i] = sum[i - 1] + a[i];
    cnt[id]++;
    if (cnt[id] >= k) {
      flag = true;
      break;
    }
  }
  if (flag) {
    puts("0");
    return 0;
  }
  for (register long long i = 1; i <= n; ++i) {
    long long lv, rv, id = Getid(a[i]);
    lv = a[i] * (l[id] - 1) - sum[l[id] - 1];
    rv = (sum[n] - sum[r[id]]) - a[i] * (n - r[id]);
    k -= cnt[id];
    if (l[id] - 1 >= k) ans = min(ans, lv - (l[id] - 1 - k));
    if (n - r[id] >= k) ans = min(ans, rv - (n - r[id] - k));
    ans = min(ans, lv + rv - (n - cnt[id] - k));
    k += cnt[id];
  }
  cout << ans << endl;
  return 0;
}