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;
long long fac[2000005];
void pre() {
  fac[0] = 1;
  fac[1] = 1;
  for (int i = 2; i < 200005; i++) fac[i] = (i * fac[i - 1]) % 1000000007;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return res % 1000000007;
}
long long ncr(long long n, long long r) {
  return ((fac[n] * power(fac[r], 1000000007 - 2)) % 1000000007 *
          power(fac[n - r], 1000000007 - 2)) %
         1000000007;
}
long long n, m;
long long a[200005];
long long tot = 0;
long long can(long long p6) {
  long long sum1 = 0;
  long long sum2 = 0;
  long long f = 0;
  for (int i = 0; i < n; i++) {
    long long have = 0;
    long long p1 = tot - sum1 - a[i];
    long long p2 = i * a[i] - sum1;
    have += p2;
    if (i + 1 >= m && have - (i + 1 - m) <= p6) {
      f = 1;
      break;
    }
    long long p3 = (n - i - 1) * a[i];
    p1 -= p3;
    if (n - i >= m && p1 - (n - i - m) <= p6) {
      f = 1;
      break;
    }
    have += p1;
    long long can1 = n - m;
    if (have - can1 <= p6) {
      f = 1;
      break;
    }
    sum1 += a[i];
  }
  return f;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    cin >> n >> m;
    map<long long, long long> mp;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      mp[a[i]]++;
      tot += a[i];
    }
    sort(a, a + n);
    for (auto it : mp) {
      if (it.second >= m) {
        cout << "0" << endl;
        return 0;
      }
    }
    long long lo = 1;
    long long hi = 1e18;
    long long ans = -1;
    while (lo <= hi) {
      long long mid = (lo + hi) / 2;
      if (can(mid)) {
        ans = mid;
        hi = mid - 1;
      } else {
        lo = mid + 1;
      }
    }
    cout << ans << endl;
  }
}