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;
inline void normal(long long &a) { a = (a + 1000000007) % 1000000007; }
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInv(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInv(b));
}
vector<long long> val[200001];
void km() {
  long long n, k;
  cin >> n >> k;
  map<long long, long long> mp;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  sort((v).begin(), (v).end());
  for (long long i = 0; i < n; i++) {
    if (mp[v[i]] >= k) {
      cout << 0 << '\n';
      return;
    }
  }
  for (long long i = 0; i < n; i++) {
    long long cnt = 0;
    long long x = v[i];
    while (x > 0) {
      val[x].push_back(cnt);
      x /= 2;
      cnt++;
    }
  }
  long long ans = INT_MAX;
  for (long long i = 1; i < 200001; i++) {
    if (val[i].size() < k) continue;
    sort((val[i]).begin(), (val[i]).end());
    long long zero = 0;
    ans = min(ans, accumulate(val[i].begin(), val[i].begin() + k, zero));
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(18);
  ;
  long long t;
  t = 1;
  while (t--) {
    km();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}