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 q, n;
long long presum[200005], precant[200005];
map<long long, long long> freq;
vector<long long> ar;
long long cost(int id) {
  long long k = ar[id];
  pair<long long, long long> may, men;
  men.first = precant[id];
  men.second = (k - 1) * precant[id] - presum[id];
  may.first = precant[n] - precant[id + 1];
  may.second = presum[n] - presum[id + 1] - (k + 1) * may.first;
  if (men.first >= q && may.first >= q) {
    return min(may.second, men.second) + q;
  } else if (men.first >= q) {
    return men.second + q;
  } else if (may.first >= q) {
    return may.second + q;
  } else {
    return men.second + may.second + q;
  }
}
int main() {
  cin >> n >> q;
  ar.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    freq[ar[i]]++;
  }
  sort(ar.begin(), ar.end());
  ar.resize(unique(ar.begin(), ar.end()) - ar.begin());
  n = ar.size();
  presum[0] = 0;
  precant[0] = 0;
  for (int i = 1; i <= n; i++) {
    presum[i] = presum[i - 1] + ar[i - 1] * freq[ar[i - 1]];
    precant[i] = precant[i - 1] + freq[ar[i - 1]];
  }
  long long mini = 1e18;
  for (int i = 0; i < n; i++) {
    q -= freq[ar[i]];
    if (q <= 0) {
      mini = 0;
      continue;
    }
    mini = min(mini, cost(i));
    q += freq[ar[i]];
  }
  cout << mini << endl;
  return 0;
}