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>
const long long int mod = 9999999999999983;
long long int primes[6] = {1125899906842597, 1495921043, 1005985879,
                           1495921043,       1005985879, 1495921043};
using namespace std;
vector<long long int> adj[2000009];
long long int parent[2000009];
long long int vis[2000009];
long long int level[2000009];
long long int dist[2000009];
long long int dp[100009];
long long int arr[100009];
long long int brr[2000009];
long long int crr[2000009];
long long int hashing[2000009];
long long int ar[509][509];
long long int br[509][509];
long long int cr[509][509];
long long int multiply(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int sub(long long int a, long long int b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res;
}
int main() {
  int start_s = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  vector<long long int> vect, vt, vct;
  vector<long long int>::iterator itt;
  set<long long int> st;
  set<long long int>::iterator itr;
  map<long long int, long long int> mp, mpp, mp1, mp2;
  map<long long int, long long int>::iterator it;
  long long int i, j, k, n, m, p, res1, res2, q, t, a, b, c,
      maxi = INT_MIN, mini = INT_MAX, sum = 0, ans = 0, res = 0, val = 0,
      ans1 = 0, ans2 = 0, rem = 0, diff = 0, cnt = 0, l, r, flag = 0, e, index,
      val1 = 0, val2 = 0, z, h = 0, u, v, mid, len, tot, fl = 0;
  string str, str1, str2;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  sort(arr + 1, arr + n + 1);
  for (i = 1; i <= n; i++) {
    cnt = 0;
    mp[arr[i]]++;
    if (mp[arr[i]] >= k) {
      vis[arr[i]] = 1;
    }
    val = arr[i];
    while (val > 0) {
      cnt++;
      val /= 2;
      if (vis[val] == 0) {
        mp[val]++;
        mpp[val] += cnt;
        if (mp[val] >= k) {
          vis[val] = 1;
        }
      }
    }
  }
  for (it = mp.begin(); it != mp.end(); it++) {
    if (it->second >= k) mini = min(mini, mpp[it->first]);
  }
  cout << mini;
}