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;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 10;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  map<int, long long> mp;
  map<int, long long> posf;
  map<int, long long> posb;
  int n, m, q, k;
  cin >> n >> m;
  long long pre[N], suf[N];
  long long a[N];
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  a[0] = 0;
  a[n + 1] = 0;
  pre[0] = 0;
  suf[n + 1] = 0;
  for (int i = 1; i <= n; i++) {
    mp[a[i]]++;
    if (a[i] != a[i - 1]) posf[a[i]] = i - 1;
    if (mp[a[i]] >= m) {
      cout << 0 << '\n';
      return 0;
    }
    pre[i] = pre[i - 1] + a[i];
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] != a[i + 1]) posb[a[i]] = n - i;
    suf[i] = suf[i + 1] + a[i];
  }
  long long ans = 2e15;
  for (auto x : mp) {
    int need = m - x.second;
    if (posf[x.first] >= need) {
      long long t = (x.first - 1) * (posf[x.first]) - pre[posf[x.first]] + need;
      ans = (ans > t ? t : ans);
    }
    if (posb[x.first] >= need) {
      long long t =
          suf[n - posb[x.first] + 1] - (x.first + 1) * (posb[x.first]) + need;
      ans = (ans > t ? t : ans);
    }
    long long t = suf[n - posb[x.first] + 1] - (x.first + 1) * (posb[x.first]) +
                  need + (x.first - 1) * (posf[x.first]) - pre[posf[x.first]];
    ans = (ans > t ? t : ans);
  }
  cout << ans << '\n';
}