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
#include <bits/stdc++.h>
#pragma GCC optimization("O2")
#pragma GCC optimization("avx2, unroll-loops")
using namespace std;
const int N = 2e5 + 5;
int n, k, a[N], ans;
vector<int> V[N];
void read_input() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
}
void add(int x) {
  int cnt = 0;
  while (x) {
    V[x].push_back(cnt++);
    x >>= 1;
  }
  V[x].push_back(cnt);
}
void solve() {
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    add(a[i]);
  }
  ans = 1e9;
  for (int i = 0; i < N; i++) {
    if ((int)V[i].size() >= k) {
      ans = min(ans, accumulate(V[i].begin(), V[i].begin() + k, 0));
    }
  }
}
void write_output() { cout << ans; }
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) {
    read_input(), solve(), write_output();
  }
  return 0;
}
