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;
int mod = 998244353;
long long nop = 100;
vector<int> primes(nop, 1);
void sieve() {
  primes[1] = primes[0] = 0;
  for (long long i = 2; i * i <= nop; i++) {
    if (primes[i]) {
      for (long long j = i * i; j <= nop; j += i) {
        primes[j] = 0;
      }
    }
  }
}
long long gcd(long long a, long long b) {
  if (b > a) swap(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
void extgcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = a;
    y = b;
    return;
  }
  extgcd(b % a, a, x, y);
  long long y1 = y, x1 = x;
  x = y1 - (b / a) * x1;
  y = x1;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long bexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans *= num;
    }
    num *= num;
    e >>= 1;
  }
  return ans;
}
long long mexp(long long num, long long e) {
  long long ans = 1;
  while (e > 0) {
    if (e & 1) {
      ans = (ans * num) % mod;
    }
    num = (num * num) % mod;
    e >>= 1;
  }
  return ans % mod;
}
long long modinv(long long a) { return bexp(a, mod - 2); }
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  };
  sort(a.begin(), a.end());
  vector<pair<long long, long long> > info(200001,
                                           pair<long long, long long>(0, 0));
  for (long long i = 0; i < n; i++) {
    info[a[i]].first++;
    long long cnt = 1;
    while (a[i]) {
      a[i] >>= 1;
      info[a[i]].first++;
      if (info[a[i]].first <= k) info[a[i]].second += (cnt++);
    }
  }
  long long ans = 0x3f3f3f3f;
  for (long long i = 0; i < 200001; i++) {
    if (info[i].first >= k) {
      ans = min(ans, info[i].second);
    }
  }
  cout << ans << endl;
  return 0;
}