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 int pw(long long int a, long long int p,
                 long long int m = 1000000007) {
  if (p == 1) return a;
  long long int t = pw(a, p / 2);
  t = (m + (t * t) % m) % m;
  if (p % 2 != 0) t *= a;
  return (t + m) % m;
}
long long int nts[200005][22], n, k;
long long int getk(long long int p) {
  long long int ms = 0;
  for (long long int i = 0; i < (21); i++) ms += nts[p][i];
  if (ms < k) return 1000000007;
  long long int ta = 0, tc = 0;
  for (long long int i = 0; i < (21); i++) {
    if (ta + nts[p][i] >= k) {
      long long int req = k - ta;
      ta += req;
      tc += req * i;
    } else {
      ta += nts[p][i];
      tc += nts[p][i] * i;
    }
  }
  return tc;
}
void solve() {
  cin >> n >> k;
  for (long long int i = 0; i < (n); i++) {
    long long int a;
    cin >> a;
    long long int steps = 0;
    while (a > 0) {
      nts[a][steps]++;
      a = a / 2;
      steps++;
    }
    nts[a][steps]++;
  }
  long long int mi = 1000000007;
  for (long long int i = 0; i < (200005); i++) mi = min(mi, getk(i));
  cout << mi << endl;
}
int main() {
  ios::sync_with_stdio(0);
  long long int t = 1;
  for (long long int i = 0; i < (t); i++) solve();
}