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
#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  long long solve(int n, int k, vector<int>& A) {
    int N = *max_element(A.begin(), A.end());
    vector<vector<int>> S(N + 1);
    for (auto x : A) {
      int h = 0;
      int bk = x;
      while (x > 0) {
        ++h;
        x >>= 1;
      }
      x = bk;
      S[x].push_back(h);
    }
    auto merge = [&](vector<int>& A, vector<int>& B, vector<int>& C) {
      int n = A.size(), m = B.size();
      C.resize(n + m);
      for (int i = 0, j = 0, k = 0; i < n || j < m; ++k) {
        if (i == n) {
          C[k] = B[j++];
          continue;
        }
        if (j == m) {
          C[k] = A[i++];
          continue;
        }
        if (A[i] <= B[j]) {
          C[k] = A[i++];
        } else {
          C[k] = B[j++];
        }
      }
    };
    long long res = 1LL << 30;
    vector<int> tmp;
    for (int i = N; i > 0; --i) {
      int h = 0;
      int x = i;
      while (x > 0) {
        ++h;
        x >>= 1;
      }
      x = i;
      int l = x << 1, r = x << 1 | 1;
      if (l <= N) {
        if (r <= N) {
          merge(S[l], S[r], tmp);
        } else {
          tmp = S[l];
        }
      } else {
        tmp.clear();
      }
      for (auto x : tmp) {
        S[i].push_back(x);
      }
      if (S[i].size() >= k) {
        long long cand = 0;
        for (int j = 0; j < k; ++j) {
          cand += S[i][j] - h;
        }
        res = min(res, cand);
      }
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int n, k;
  cin >> n >> k;
  vector<int> A(n);
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
  }
  cout << sol.solve(n, k, A) << "\n";
  return 0;
}
