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 solve(const vector<long long> &arr) {
  vector<pair<long long, long long>> cnts;
  {
    map<long long, long long> temp;
    for (long long i = 0; i < n; ++i) {
      ++temp[arr[i]];
    }
    for (auto &p : temp) {
      cnts.push_back(p);
    }
  }
  for (auto &p : cnts) {
    if (p.second >= k) {
      return 0;
    }
  }
  vector<long long> suffixSum(cnts.size() + 1, 0);
  for (long long i = (long long)cnts.size() - 1; i >= 0; --i) {
    suffixSum[i] = suffixSum[i + 1] + cnts[i].second;
  }
  vector<long long> suffix(cnts.size(), 0);
  for (long long i = (long long)cnts.size() - 2; i >= 0; --i) {
    assert(cnts[i + 1].first > cnts[i].first);
    suffix[i] =
        suffix[i + 1] + (cnts[i + 1].first - cnts[i].first) * suffixSum[i + 1];
  }
  long long ops = 0;
  long long prefixSum = 0;
  long long res = 1e18;
  for (long long i = 0; i < cnts.size(); ++i) {
    if (prefixSum + cnts[i].second >= k) {
      assert(prefixSum >= k - cnts[i].second);
      res = min(res, ops + (cnts[i].first - 1 - cnts[i - 1].first) * prefixSum +
                         (k - cnts[i].second));
      break;
    }
    if (i) {
      ops += (cnts[i].first - cnts[i - 1].first) * prefixSum;
    }
    prefixSum += cnts[i].second;
    res = min(res,
              ops + suffix[i + 1] +
                  suffixSum[i + 1] * (cnts[i + 1].first - 1 - cnts[i].first) +
                  (k - prefixSum));
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  vector<long long> arr(n);
  vector<long long> arr2(n);
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i];
    arr2[i] = -arr[i];
  }
  cout << min(solve(arr), solve(arr2));
}