There is a toy building consisting of n towers. Each tower consists of several cubes standing on each other. The i-th tower consists of h_i cubes, so it has height h_i.

Let's define operation slice on some height H as following: for each tower i, if its height is greater than H, then remove some top cubes to make tower's height equal to H. Cost of one "slice" equals to the total number of removed cubes from all towers.

Let's name slice as good one if its cost is lower or equal to k (k ≥ n).

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Calculate the minimum number of good slices you have to do to make all towers have the same height. Of course, it is always possible to make it so.

Input

The first line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5, n ≤ k ≤ 10^9) — the number of towers and the restriction on slices, respectively.

The second line contains n space separated integers h_1, h_2, ..., h_n (1 ≤ h_i ≤ 2 ⋅ 10^5) — the initial heights of towers.

Output

Print one integer — the minimum number of good slices you have to do to make all towers have the same heigth.

Examples

Input

5 5
3 1 2 2 4


Output

2


Input

4 5
2 3 4 5


Output

2

Note

In the first example it's optimal to make 2 slices. The first slice is on height 2 (its cost is 3), and the second one is on height 1 (its cost is 4).
#include <bits/stdc++.h>
using namespace std;
long long int M = 1e9 + 7;
long long int n, k;
long long int x, y, z, t, q, u, a1, l, r;
long long int a[300007];
long long int vis[100001];
long long int b[100007];
map<long long int, long long int> dp;
long long int dp1[507][507];
long long int sum[100007][2];
long long int s[5007];
pair<long long int, long long int> p[27];
pair<long long int, long long int> p1[27];
bool check(long long int j, long long int i) {
  if ((p[i].second <= p[j].first) || (p[i].first >= p[j].second)) return true;
  return false;
}
long long int m, d;
vector<pair<long long int, long long int> > v[100007];
vector<long long int> v1[2001];
long long int m1[20000007];
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  if (a.first == b.first) return 1;
  return a.first > b.first;
}
map<long long int, long long int> c;
long long int root;
vector<long long int> ans;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int mx = 1e10;
long long int solve(long long int i, long long int p) {
  if (i > mx) return i;
  if (p == 100) return i;
  if (dp.find(i) != dp.end()) return dp[i];
  long long int i1 = i;
  long long int s = 0;
  long long int s1 = 0;
  while (i1 > 0) {
    s += i1 % 10;
    i1 /= 10;
    s1++;
  }
  if (s1 == 1)
    dp[i] = min(solve(i + d, p + 1), s);
  else
    dp[i] = min(solve(s, p + 1), solve(i + d, p + 1));
  cout << dp[i] << " " << i << " " << p << endl;
  return dp[i];
}
const long long int mi2 = 2e7 + 5;
bitset<mi2> chk;
int main() {
  cin >> n >> k;
  map<long long int, long long int> m, m1;
  long long int mx = -1;
  long long int c = 0;
  long long int c4 = 1000000007;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    a[i] = x;
    m[x]++;
    mx = max(mx, x);
    c4 = min(c4, x);
  }
  if (c4 == mx) return cout << 0, 0;
  c = m[mx];
  for (int i = mx - 1; i >= 0; i--) {
    if (m[i + 1] >= 0) {
      long long int c1 = c;
      c += m[i];
      m[i] += m[i + 1] + c1;
    }
  }
  for (int i = mx - 1; i >= 0; i--) {
    m1[i] = m[i + 1];
  }
  long long int c6 = 0;
  long long int k1 = k;
  long long int c7 = -1;
  for (int i = mx; i >= 1; i--) {
    if (k >= m1[i - 1]) {
      continue;
    }
    if (i <= c4) {
      break;
    }
    c7 = i;
    k = k1;
    k += m1[i];
    c6++;
  }
  if (c7 != c4) c6++;
  cout << c6;
}
