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;
using namespace std;
const long long mod2 = 998244353;
long long spf[100005];
long long fpow(long long x, long long y) {
  x = x % 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= 1000000007;
    y = y >> 1;
    x = x * x;
    x %= 1000000007;
  }
  return res;
}
long long inv(long long a, long long m = 1000000007) {
  long long c = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += c;
  return x;
}
long long cnt[200009];
long long res[200009];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    long long x = a[i];
    long long moves = 0;
    while (x) {
      if (cnt[x] < k) {
        cnt[x]++;
        res[x] += moves;
      }
      x /= 2;
      moves++;
    }
    if (cnt[0] < k) {
      cnt[0]++;
      res[0] += moves;
    }
  }
  long long result = LLONG_MAX;
  for (int i = 0; i <= 200005; i++) {
    if (cnt[i] >= k) result = min(result, res[i]);
  }
  cout << result;
}