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;
int n, k, t;
struct Stats {
  long long int nBefore, nAfter, sLeft, sRight;
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  map<int, int> num;
  for (int i = 0; i < n; i++) {
    cin >> t;
    if (num[t]) {
      num[t]++;
    } else
      num[t] = 1;
  }
  t = num.size();
  for (auto a : num) {
    if (a.second >= k) {
      cout << 0 << '\n';
      return 0;
    }
  }
  vector<pair<int, int>> v(num.begin(), num.end());
  vector<Stats> s(t);
  long long int rsum = v[t - 1].second, lsum = v[0].second, rsteps = 0,
                lsteps = 0;
  s[0].nBefore = lsum;
  s[0].sLeft = 0;
  s[t - 1].nAfter = rsum;
  s[t - 1].sRight = 0;
  for (int i = 1, j = t - 2; i < t; ++i, --j) {
    lsteps += (v[i].first - v[i - 1].first) * lsum;
    lsum += v[i].second;
    rsteps += (v[j + 1].first - v[j].first) * rsum;
    rsum += v[j].second;
    s[i].nBefore = lsum;
    s[i].sLeft = lsteps;
    s[j].nAfter = rsum;
    s[j].sRight = rsteps;
  }
  long long int min_moves = LONG_LONG_MAX;
  for (auto it = s.begin(); it != s.end(); ++it) {
    long long int m;
    bool cond1 = it->nBefore >= k, cond2 = it->nAfter >= k;
    if (cond1 && cond2) {
      m = min(it->sLeft - it->nBefore, it->sRight - it->nAfter) + k;
    } else if (cond1) {
      m = it->sLeft - it->nBefore + k;
    } else if (cond2) {
      m = it->sRight - it->nAfter + k;
    } else {
      m = it->sLeft + it->sRight - n + k;
    }
    min_moves = min(min_moves, m);
  }
  cout << min_moves << '\n';
  return 0;
}