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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool mod(long double a, long double b) { return a / b - floor(a / b); }
long long power(long long first, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = result * first;
    first = first * first;
    n = n / 2;
  }
  return result;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  unordered_map<long long, long long> m;
  long long count[200002] = {0};
  vector<long long> v(n);
  for (long long &first : v) cin >> first;
  long long mn = INT_MAX;
  sort((v).begin(), (v).end());
  for (long long i = 0; i < n; i++) {
    long long flag = 0;
    long long first = v[i];
    m[first]++;
    if (m[first] == k) mn = 0;
    while (first > 1) {
      first /= 2;
      flag++;
      m[first]++;
      count[first] += flag;
      if (m[first] == k) mn = min(mn, count[first]);
    }
  }
  cout << mn;
}