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 + 5;
int n, k;
long long a[N], tot[N], sum[N], c[N], d[N];
vector<long long> b;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b.push_back(a[i]);
  }
  sort((b).begin(), (b).end());
  b.resize(unique((b).begin(), (b).end()) - b.begin());
  for (int i = 1; i <= n; i++)
    ++tot[lower_bound((b).begin(), (b).end(), a[i]) - b.begin()];
  sum[0] = tot[0];
  for (int i = 1; i < (int)(b).size(); i++) sum[i] = tot[i] + sum[i - 1];
  for (int i = 1; i < (int)(b).size(); i++)
    c[i] = c[i - 1] + sum[i - 1] * (b[i] - b[i - 1]);
  for (int i = (int)(b).size() - 2; i >= 0; i--)
    d[i] = d[i + 1] + (sum[(int)(b).size() - 1] - sum[i]) * (b[i + 1] - b[i]);
  long long mn = 0x3f3f3f3f3f3f3f3f;
  for (int i = 0; i < (int)(b).size(); i++) {
    if (tot[i] >= k) {
      puts("0");
      return 0;
    }
    long long sum1 = 0, sum2 = 0;
    if (i - 1 >= 0) sum1 = c[i - 1] + sum[i - 1] * (b[i] - b[i - 1] - 1);
    if (i + 1 < (int)(b).size())
      sum2 = d[i + 1] +
             (sum[(int)(b).size() - 1] - sum[i]) * (b[i + 1] - b[i] - 1);
    if (i - 1 >= 0 && sum[i - 1] + tot[i] >= k) mn = min(mn, sum1 + k - tot[i]);
    if (sum[(int)(b).size() - 1] - sum[i] + tot[i] >= k)
      mn = min(mn, sum2 + k - tot[i]);
    mn = min(mn, sum1 + sum2 + k - tot[i]);
  }
  printf("%lld\n", mn);
  return 0;
}