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 double PI = acos(-1);
const int N = 1e6 + 7;
long long a[N], spre[N], ssuf[N], npre[N], nsuf[N];
map<int, int> num;
int k, n;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i], num[a[i]]++;
  sort(a + 1, a + n + 1);
  n = unique(a + 1, a + n + 1) - a - 1;
  for (int i = 1; i <= n; i++)
    npre[i] = npre[i - 1] + num[a[i]], spre[i] = spre[i - 1] + a[i] * num[a[i]];
  for (int i = n; i >= 1; i--)
    nsuf[i] = nsuf[i + 1] + num[a[i]], ssuf[i] = ssuf[i + 1] + a[i] * num[a[i]];
  long long ans = 1e18 + 7;
  for (int i = 1; i <= n; i++) {
    int m = k - npre[i] + npre[i - 1];
    if (m <= 0) {
      ans = 0;
      continue;
    }
    long long tmp1 = npre[i - 1] * (a[i] - 1) - spre[i - 1];
    long long tmp2 = ssuf[i + 1] - nsuf[i + 1] * (a[i] + 1);
    if (npre[i - 1] >= m) {
      ans = min(tmp1 + m, ans);
    }
    if (nsuf[i + 1] >= m) {
      ans = min(tmp2 + m, ans);
    }
    ans = min(tmp1 + tmp2 + m, ans);
  }
  cout << max(ans, 0ll) << endl;
}