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>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-9;
const long long MOD = 1e9 + 7;
const long long INF = 1e18 + 7;
const long long MAXN = 2e5 + 7;
long long n, k;
vector<long long> g[MAXN];
void process(long long m) {
  int i = 0;
  for (; m; i++) {
    g[m].push_back(i);
    m >>= 1;
  }
  g[m].push_back(i);
}
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    long long m;
    cin >> m;
    process(m);
  }
  long long ans = 1e18;
  for (int i = 0; i <= 200000; i++) {
    int sz = g[i].size();
    if (sz < k) continue;
    sort(g[i].begin(), g[i].end());
    long long res = 0;
    for (int j = 0; j < k; j++) {
      res += g[i][j];
    }
    ans = min(ans, res);
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int i, Case = 1;
  while (Case--) solve();
  return 0;
}
