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 ll = int64_t;
int main() {
  using namespace std;
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  vector<vector<int>> dp(n, vector<int>(200010, -1));
  int x, maxX = 0;
  for (int i = 0, iLen = (n); i < iLen; ++i) {
    cin >> x;
    maxX = max(maxX, x);
    a[i] = x;
    dp[i][x] = 0;
    while (true) {
      dp[i][x / 2] = dp[i][x] + 1;
      x /= 2;
      if (x == 0) break;
    }
  }
  int ans = 10000000;
  for (int i = 0, iLen = (maxX + 1); i < iLen; ++i) {
    auto hoge = vector<int>();
    for (int j = 0, jLen = (n); j < jLen; ++j)
      if (dp[j][i] > -1) hoge.emplace_back(dp[j][i]);
    if (hoge.size() < k) continue;
    sort(hoge.begin(), hoge.end());
    int total = 0;
    for (int j = 0, jLen = (k); j < jLen; ++j) total += hoge[j];
    ans = min(ans, total);
  }
  cout << ans << endl;
  return 0;
}