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, k;
long long a[200050];
long long pre[200050];
long long suf[200050];
map<long long, long long> cnt;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + a[i];
  }
  for (int i = n; i >= 1; i--) {
    suf[i] = suf[i + 1] + a[i];
  }
  long long mi = 1e18;
  for (int i = 1; i <= n; i++) {
    if (cnt[a[i]] >= k) {
      cout << 0 << endl;
      return 0;
    }
    long long need = k - cnt[a[i]];
    long long cost1 = (a[i] - 1) * (i - 1) - pre[i - 1];
    long long cost2 = suf[i + cnt[a[i]]] - (a[i] + 1) * (n - i - cnt[a[i]] + 1);
    long long L = i - 1;
    long long R = n - i - cnt[a[i]] + 1;
    if (L >= need && R >= need)
      mi = min(mi, min(cost1, cost2) + need);
    else if (L >= need && R < need)
      mi = min(mi, cost1 + need);
    else if (L < need && R >= need)
      mi = min(mi, cost2 + need);
    else {
      mi = min(cost1 + cost2 + need, mi);
    }
    i += cnt[a[i]] - 1;
  }
  cout << mi << endl;
}