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;
long long n, m, A[1000009];
vector<long long> v[200009];
int main() {
  long long tc = 1, t = 0, i, j, k, a, b, c, d, s;
  while (t++ < tc) {
    cin >> n >> m;
    for (i = 1; i <= n; i++) {
      cin >> A[i];
      k = 0;
      while (A[i]) {
        v[A[i]].push_back(k);
        k++;
        A[i] /= 2;
      }
    }
    long long ans = LLONG_MAX;
    for (i = 0; i < 200009; i++) {
      if (v[i].size() >= m) {
        sort(v[i].begin(), v[i].end());
        long long cur = 0;
        for (j = 0; j < m; j++) cur += v[i][j];
        ans = min(ans, cur);
      }
    }
    cout << ans;
  }
  return 0;
}