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;
mt19937_64 mt(time(0));
long long solve(long long target, int k, long long sum1, long long sum2,
                int cnt1, int cnt2) {
  if (k <= 0) return 0;
  long long ret1 = (target - 1) * cnt1 - sum1;
  ret1 += min(k, cnt1);
  int k1 = k - min(k, cnt1);
  if (k1 > 0) {
    ret1 += sum2 - (target + 1) * cnt2;
    ret1 += k1;
  }
  long long ret2 = sum2 - (target + 1) * cnt2;
  ret2 += min(k, cnt2);
  int k2 = k - min(k, cnt2);
  if (k2 > 0) {
    ret2 += (target - 1) * cnt1 - sum1;
    ret2 += k2;
  }
  return min(ret1, ret2);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  int n, k, val;
  cin >> n >> k;
  map<int, int> table;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> val;
    table[val]++;
    sum += val;
  }
  int cur_cnt = 0;
  long long cur_sum = 0, ret = LLONG_MAX;
  for (const auto& item : table) {
    int tmp_cnt = item.second;
    long long tmp_sum = item.first * (long long)tmp_cnt;
    long long tmp =
        solve(item.first, k - tmp_cnt, cur_sum, sum - cur_sum - tmp_sum,
              cur_cnt, n - cur_cnt - tmp_cnt);
    ret = min(ret, tmp);
    cur_sum += tmp_sum;
    cur_cnt += tmp_cnt;
  }
  cout << ret << '\n';
  return 0;
}