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 md = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ll n, k;
  cin >> n >> k;
  vector<ll> a(n);
  for (ll i = 0; i < n; i++) {
    cin >> a[i];
  }
  map<ll, priority_queue<ll>> mp;
  map<ll, ll> sum;
  for (ll i = 0; i < n; i++) {
    ll cur = a[i];
    ll op = 0;
    while (1) {
      if (mp[cur].size() < k) {
        mp[cur].push(op);
        sum[cur] += op;
      } else {
        ll tp = mp[cur].top();
        if (tp > op) {
          sum[cur] -= tp;
          sum[cur] += op;
          mp[cur].pop();
          mp[cur].push(op);
        }
      }
      cur >>= 1;
      op++;
      if (cur == 0) {
        if (mp[cur].size() < k) {
          mp[cur].push(op);
          sum[cur] += op;
        } else {
          ll tp = mp[cur].top();
          if (tp > op) {
            sum[cur] -= tp;
            sum[cur] += op;
            mp[cur].pop();
            mp[cur].push(op);
          }
        }
        break;
      }
    }
  }
  ll ans = 1e18;
  for (auto i : mp) {
    priority_queue<ll> pq = i.second;
    if ((ll)pq.size() < k) {
      continue;
    }
    ans = min(ans, sum[i.first]);
  }
  cout << ans << '\n';
  return 0;
}