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
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 10, inf = 0x3f3f3f3f, mod = 1000000007;
void read() {}
template <typename T, typename... T2>
inline void read(T &x, T2 &...oth) {
  x = 0;
  long long ch = getchar(), f = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  if (f) x = -x;
  read(oth...);
}
vector<long long> vec;
long long id(long long x) {
  return lower_bound(vec.begin(), vec.end(), x) - vec.begin() + 1;
}
long long sum[maxn], a[maxn], pre[maxn], suf[maxn], cnt[maxn], pcnt[maxn],
    scnt[maxn];
signed main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, k;
  cin >> n >> k;
  map<long long, long long> mp;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]]++;
    if (mp[a[i]] >= k) {
      cout << 0 << endl;
      exit(0);
    }
  }
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + a[i];
    vec.push_back(a[i]);
  }
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  long long rec = 0;
  for (long long i = 1; i <= n; i++) {
    rec += a[i];
    long long now = id(a[i]);
    cnt[now]++;
    if (a[i] != a[i + 1]) {
      pre[now] = (a[i] - 1) * (i - cnt[now]) - (rec - cnt[now] * a[i]);
      pcnt[now] = pcnt[now - 1] + cnt[now - 1];
    }
  }
  rec = 0;
  for (long long i = n; i >= 1; i--) {
    rec += a[i];
    if (a[i - 1] != a[i]) {
      long long now = id(a[i]);
      suf[now] = (rec - cnt[now] * a[i]) - (n - i + 1 - cnt[now]) * (a[i] + 1);
      scnt[now] = scnt[now + 1] + cnt[now + 1];
    }
  }
  long long ans = LLONG_MAX;
  ans = min(ans, a[k] * k - sum[k]);
  ans = min(ans, (sum[n] - sum[n - k]) - a[n - k + 1] * k);
  for (auto &x : vec) {
    long long now = id(x);
    long long res = k - cnt[now];
    ans = min(ans, pre[now] + suf[now] + res);
    if (pcnt[now] >= res) ans = min(ans, pre[now] + res);
    if (scnt[now] >= res) ans = min(ans, suf[now] + res);
  }
  cout << ans << endl;
  return 0;
}
