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;
template <class T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
void setmax(T& a, T b) {
  a = max(a, b);
};
template <typename T>
void setmin(T& a, T b) {
  a = min(a, b);
};
template <typename T>
bool in(T v, T lo, T hi) {
  return lo <= v && v <= hi;
};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 10;
int n, k;
int a[N];
map<int, long long> lcost;
map<int, long long> rcost;
map<int, int> lnum;
map<int, int> rnum;
map<int, int> cnt;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int num = 0;
  long long cost = 0;
  int at = 0;
  vector<pair<int, int>> rev;
  for (auto& [x, mult] : cnt) {
    rev.push_back({x, mult});
    cost += ((x - 1) - at) * 1ll * num;
    at = x - 1;
    lcost[x] = cost;
    lnum[x] = num;
    cost += num;
    at = x;
    num += mult;
  }
  reverse(rev.begin(), rev.end());
  {
    int num = 0;
    long long cost = 0;
    int at = 1e9 + 5;
    for (auto& [x, mult] : rev) {
      cost += (at - (x + 1)) * 1ll * num;
      at = x + 1;
      rcost[x] = cost;
      rnum[x] = num;
      cost += num;
      at = x;
      num += mult;
    }
  }
  long long ans = 1e18;
  for (auto& [x, mult] : cnt) {
    if (mult >= k) {
      setmin(ans, 0ll);
    }
    if (mult + rnum[x] >= k) {
      setmin(ans, rcost[x] + max(0, k - mult));
    }
    if (mult + lnum[x] >= k) {
      setmin(ans, lcost[x] + max(0, k - mult));
    }
    setmin(ans, lcost[x] + rcost[x] + max(0, k - mult));
  }
  cout << ans << '\n';
  return 0;
}