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;
vector<long long int> vec;
set<long long int> s;
long long int cnt[200009], val[200009];
long long int cnt2[200009], val2[200009];
long long int n, k;
map<long long int, long long int> mp;
long long int call(long long int x) {
  long long int a, b, c, y, i, j, ans;
  long long int ans1, ans2;
  ans1 = ans2 = 0;
  ans = 0;
  if (mp[x] >= k) {
    return ans;
  }
  y = k - mp[x];
  a = lower_bound(vec.begin(), vec.end(), x) - vec.begin();
  a--;
  b = upper_bound(vec.begin(), vec.end(), x) - vec.begin();
  if (b == n) {
    ans += val[a] + (((x - 1) - vec[a]) * cnt[a]) + y;
  } else if (a < 0) {
    ans += val2[b] + ((vec[b] - (x + 1)) * cnt2[b]) + y;
  } else {
    long long int yy = y;
    ans1 += val[a];
    c = min(cnt[a], y);
    ans1 += (((x - 1) - vec[a]) * cnt[a]) + c;
    y -= c;
    if (y > 0) {
      ans1 += val2[b];
      ans1 += ((vec[b] - (x + 1)) * cnt2[b]) + y;
    }
    y = yy;
    ans2 += val2[b];
    c = min(cnt2[b], y);
    ans2 += ((vec[b] - (x + 1)) * cnt2[b]) + c;
    y -= c;
    if (y > 0) {
      ans2 += val[a];
      ans2 += (((x - 1) - vec[a]) * cnt[a]) + y;
    }
    ans = min(ans1, ans2);
  }
  return ans;
}
int main() {
  long long int a, b, c, d, e, i, j, l, m, x, y, t, p, r;
  scanf("%lld%lld", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a);
    mp[a]++;
    s.insert(a);
  }
  for (auto it = s.begin(); it != s.end(); it++) {
    vec.push_back(*it);
  }
  n = vec.size();
  cnt[0] = mp[vec[0]];
  cnt2[n - 1] = mp[vec[n - 1]];
  for (i = 1; i < n; i++) {
    cnt[i] = cnt[i - 1] + (mp[vec[i]]);
    val[i] = val[i - 1] + ((vec[i] - vec[i - 1]) * (cnt[i - 1]));
  }
  for (i = n - 2; i >= 0; i--) {
    cnt2[i] = cnt2[i + 1] + mp[vec[i]];
    val2[i] = val2[i + 1] + ((vec[i + 1] - vec[i]) * cnt2[i + 1]);
  }
  x = 999999999999999999;
  for (i = 0; i < n; i++) {
    y = call(vec[i]);
    x = min(x, y);
  }
  printf("%lld\n", x);
}