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 int N = 3e5;
vector<int> g[N];
map<int, int> a;
map<int, int> res[N];
int am[N];
int d[N];
void build(int x = 0) {
  int i = 2 * x, d = 2 * x + 1;
  if (i < N && i > 0) {
    g[x].push_back(i);
    build(i);
  }
  if (d < N) {
    g[x].push_back(d);
    build(d);
  }
}
void solve(int x, int l = 0) {
  for (auto y : g[x]) {
    solve(y, l + 1);
    for (auto z : res[y]) {
      res[x][z.first] += z.second;
    }
    am[x] += am[y];
  }
  if (a[x]) res[x][l] += a[x], am[x] += a[x];
  d[x] = l;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[x]++;
  }
  build();
  solve(0);
  int r = 1e9;
  for (int i = 0; i < N; i++) {
    if (am[i] >= k) {
      int qty = 0;
      int sum = 0;
      for (auto x : res[i]) {
        int dd = min(k - qty, x.second);
        qty += dd;
        sum += dd * (x.first - d[i]);
      }
      r = min(r, sum);
    }
  }
  cout << r << '\n';
  return 0;
}