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 long long LLINF = 0x3f3f3f3f3f3f3f3f;
class Item {
 public:
  long long val, cnt, sum;
  Item() { val = cnt = sum = 0LL; };
};
int n = 0, k = 0, tot = 0;
long long a[200005] = {};
Item s[200005] = {};
Item pre[200005] = {};
Item suf[200005] = {};
void init() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  long long last = 0, cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] != last) {
      if (last) {
        ++tot;
        s[tot].val = last;
        s[tot].cnt = cnt;
        s[tot].sum = last * cnt;
      }
      last = a[i];
      cnt = 1;
    } else {
      ++cnt;
    }
  }
  ++tot;
  s[tot].val = last;
  s[tot].cnt = cnt;
  s[tot].sum = last * cnt;
  for (int i = 1; i <= tot; ++i) {
    pre[i].cnt = pre[i - 1].cnt + s[i].cnt;
    pre[i].sum = pre[i - 1].sum + s[i].sum;
  }
  for (int i = tot; i >= 1; --i) {
    suf[i].cnt = suf[i + 1].cnt + s[i].cnt;
    suf[i].sum = suf[i + 1].sum + s[i].sum;
  }
}
inline long long costL(int i) {
  return (s[i].val - 1) * (pre[i - 1].cnt) - pre[i - 1].sum;
}
inline long long costR(int i) {
  return suf[i + 1].sum - (s[i].val + 1) * (suf[i + 1].cnt);
}
long long solve() {
  long long ans = LLINF;
  for (int i = 1; i <= tot; ++i) {
    if (s[i].cnt >= k) {
      ans = 0LL;
      break;
    } else {
      long long cnt = k - s[i].cnt;
      {
        long long cntL = min(cnt, pre[i - 1].cnt);
        long long cntR = max(cnt - cntL, 0LL);
        long long tmp = costL(i);
        if (cntR > 0) {
          tmp += costR(i);
        }
        ans = min(ans, tmp + cnt);
      }
      {
        long long cntR = min(cnt, suf[i + 1].cnt);
        long long cntL = max(cnt - cntR, 0LL);
        long long tmp = costR(i);
        if (cntL > 0) {
          tmp += costL(i);
        }
        ans = min(ans, tmp + cnt);
      }
    }
  }
  return ans;
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  init();
  printf("%lld\n", solve());
  return 0;
}