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 int N = 2e5 + 10;
int n, k;
int a[N];
int cnt[N];
int num[N];
long long pre[N], pcnt[N];
long long suf[N], scnt[N];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int ccnt = 0;
  int m = 0;
  for (int i = 1; i <= n; i++) {
    ccnt++;
    if (a[i] != a[i + 1]) {
      num[++m] = a[i];
      cnt[m] = ccnt;
      ccnt = 0;
    }
  }
  for (int i = 2; i <= m; i++) {
    pre[i] = pre[i - 1] + pcnt[i - 1] * (num[i] - num[i - 1]) +
             cnt[i - 1] * (num[i] - num[i - 1] - 1);
    pcnt[i] = pcnt[i - 1] + cnt[i - 1];
  }
  for (int i = m - 1; i >= 1; i--) {
    suf[i] = suf[i + 1] + scnt[i + 1] * (num[i + 1] - num[i]) +
             cnt[i + 1] * (num[i + 1] - num[i] - 1);
    scnt[i] = scnt[i + 1] + cnt[i + 1];
  }
  long long ans = 1e18 + 10;
  for (int i = 1; i <= m; i++) {
    int res = k - cnt[i];
    if (res <= 0) {
      cout << "0\n";
      return 0;
    }
    if (i > 1 && i < m) {
      ans = min(ans, pre[i] + suf[i] + res);
    }
    if (i > 1 && pcnt[i] >= res) {
      ans = min(ans, pre[i] + res);
    }
    if (i < m && scnt[i] >= res) {
      ans = min(ans, suf[i] + res);
    }
  }
  cout << ans << endl;
  return 0;
}