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
#include <bits/stdc++.h>
using namespace std;
using ll = signed long long int;
const int N = 2e5 + 10;
using mp = map<ll, ll>;
mp m;
ll n, k;
struct num {
  ll key, frq, lcost, rcost, lcount, rcount;
  inline ll calc() const {
    if (frq >= k) {
      return 0LL;
    }
    return min(calc(lcost, lcount, rcost, rcount),
               calc(rcost, rcount, lcost, lcount));
  }
  inline ll calc(ll ls, ll lsn, ll gr, ll grn) const {
    ll res = 0LL, rq = k - frq;
    res += ls;
    rq -= min(rq, lsn);
    if (rq > 0) {
      res += gr;
    }
    res += k - frq;
    return res;
  }
};
vector<num> v;
ll res = numeric_limits<ll>::max();
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cout << fixed;
  cout.precision(12);
  v.reserve(N);
  cin >> n >> k;
  for (int i = 0, t; i < n; i++) {
    cin >> t;
    m[t]++;
  }
  for (auto p : m) {
    v.push_back({p.first, p.second});
  }
  v.front().lcost = v.front().lcount = v.back().rcost = v.back().rcount = 0LL;
  for (int i = 1; i < v.size(); i++) {
    v[i].lcost = v[i - 1].lcost + (v[i].key - v[i - 1].key) * v[i - 1].lcount +
                 (v[i].key - v[i - 1].key - 1LL) * v[i - 1].frq;
    v[i].lcount = v[i - 1].lcount + v[i - 1].frq;
  }
  for (int i = v.size() - 2; i >= 0; i--) {
    v[i].rcost = v[i + 1].rcost + (v[i + 1].key - v[i].key) * v[i + 1].rcount +
                 (v[i + 1].key - v[i].key - 1LL) * v[i + 1].frq;
    v[i].rcount = v[i + 1].rcount + v[i + 1].frq;
  }
  for (auto& x : v) {
    res = min(res, x.calc());
  }
  cout << res;
}
