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 ll = long long;
const ll oo = 0x3f3f3f3f3f3f3f3fLL;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n, k;
  cin >> n >> k;
  vector<ll> a(n);
  map<ll, ll> am;
  for (ll i = 0; i < n; i++) cin >> a[i], am[a[i]]++;
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
  map<ll, pair<ll, ll>> mx, mn;
  mn[a[0]] = make_pair(0, am[a[0]]);
  for (ll i = 1; i < (ll)a.size(); i++) {
    mn[a[i]] =
        make_pair(mn[a[i - 1]].first + mn[a[i - 1]].second * (a[i] - a[i - 1]),
                  mn[a[i - 1]].second + am[a[i]]);
  }
  mx[a.back()] = make_pair(0, am[a.back()]);
  for (ll i = (ll)a.size() - 2; i >= 0; i--) {
    mx[a[i]] =
        make_pair(mx[a[i + 1]].first + mx[a[i + 1]].second * (a[i + 1] - a[i]),
                  mx[a[i + 1]].second + am[a[i]]);
  }
  ll res = oo;
  if (am[a[0]] < k) {
    res = min(res, mx[a[1]].first + (k - am[a[0]]) +
                       mx[a[1]].second * (a[1] - a[0] - 1));
  } else {
    res = 0;
  }
  if (am[a.back()] < k) {
    res = min(res, mn[a[(ll)a.size() - 2]].first + (k - am[a.back()]) +
                       mn[a[(ll)a.size() - 2]].second *
                           (a.back() - a[(ll)a.size() - 2] - 1));
  } else {
    res = 0;
  }
  for (ll goal = 1; goal < (ll)a.size() - 1; goal++) {
    ll mncost = mn[a[goal - 1]].first;
    ll mxcost = mx[a[goal + 1]].first;
    ll needed = max(k - am[a[goal]], 0LL);
    ll curcost = 0;
    if (needed > 0) {
      curcost += min(needed, mn[a[goal - 1]].second) +
                 mn[a[goal - 1]].second * (a[goal] - a[goal - 1] - 1) + mncost;
      needed -= min(needed, mn[a[goal - 1]].second);
    }
    if (needed > 0) {
      curcost += min(needed, mx[a[goal + 1]].second) +
                 mx[a[goal + 1]].second * (a[goal + 1] - a[goal] - 1) + mxcost;
      needed -= min(needed, mx[a[goal + 1]].second);
    }
    assert(needed == 0);
    res = min(res, curcost);
    needed = max(k - am[a[goal]], 0LL);
    curcost = 0;
    if (needed > 0) {
      curcost += min(needed, mx[a[goal + 1]].second) +
                 mx[a[goal + 1]].second * (a[goal + 1] - a[goal] - 1) + mxcost;
      needed -= min(needed, mx[a[goal + 1]].second);
    }
    if (needed > 0) {
      curcost += min(needed, mn[a[goal - 1]].second) +
                 mn[a[goal - 1]].second * (a[goal] - a[goal - 1] - 1) + mncost;
      needed -= min(needed, mn[a[goal - 1]].second);
    }
    assert(needed == 0);
    res = min(res, curcost);
  }
  cout << res << '\n';
}