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, m;
pair<long long, long long> a[2000111], b[2000111], c[2000111], d[2000111];
void lam() {
  cin >> n >> m;
  long long nn = n, kt = 0, bd = 1, qq = LLONG_MAX;
  for (int i = 1; i <= nn; i++) cin >> a[i].first;
  sort(a + 1, a + n + 1);
  for (int i = 2; i <= n + 1; i++)
    if (a[i].first != a[i - 1].first) {
      b[++kt].second = i - bd, bd = i;
      b[kt].first = a[i - 1].first;
    }
  for (int i = kt; i >= 1; i--) {
    c[i].first = c[i + 1].first + b[i].first * b[i].second;
    c[i].second = c[i + 1].second + b[i].second;
  }
  for (int i = 1; i <= kt; i++) {
    long long doa = m - b[i].second;
    pair<long long, long long> xx = pair<long long, long long>(
                                   (b[i].first - 1) * d[i - 1].second -
                                       d[i - 1].first,
                                   d[i - 1].second),
                               yy = pair<long long, long long>(
                                   c[i + 1].first -
                                       (b[i].first + 1) * c[i + 1].second,
                                   c[i + 1].second);
    if (xx.first > yy.first) swap(xx, yy);
    if (doa <= 0)
      qq = 0;
    else if (xx.second >= doa)
      qq = min(qq, xx.first + doa);
    else if (yy.second >= doa)
      qq = min(qq, yy.first + doa);
    else
      qq = min(qq, xx.first + yy.first + doa);
    d[i].first = d[i - 1].first + b[i].first * b[i].second;
    d[i].second = d[i - 1].second + b[i].second;
  }
  cout << qq;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  lam();
}