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 MOD = 1e9 + 7;
long long int prime[10000000];
long long int fac[200100], ifac[200100];
long long int inv[200100];
vector<long long int> primes;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool IsPrime(long long int x) {
  if (x == 1) return false;
  for (long long int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
void seive(long long int n) {
  prime[0] = prime[1] = 1;
  for (long long int i = 2; i * i <= n; i++)
    if (!prime[i]) {
      primes.push_back(i);
      for (long long int j = i * i; j <= 1e6; j += i) prime[j] = 1;
    }
}
bool sortMa7boub(const pair<long double, int>& a,
                 const pair<long double, int>& b) {
  if (a.first == b.first) return (a.second < b.second);
  return (a.first > b.first);
}
bool IsVowel(char c) {
  return (c == 'a' || c == 'e' || c == 'i' || c == 'u' || c == 'o' ||
          c == 'y' || c == 'A' || c == 'E' || c == 'I' || c == 'U' ||
          c == 'O' || c == 'Y');
}
bool IsEqual(string s1, string s2) {
  if (s1.length() != s2.length()) return 0;
  for (int i = 0; i < s1.length(); i++)
    if (s1[i] != s2[i]) return 0;
  return 1;
}
long long int powMod(long long int base, long long int pw) {
  long long int ret = 1;
  while (pw) {
    if (pw & 1) ret = (ret * base) % MOD;
    base = (base * base) % MOD;
    pw >>= 1;
  }
  return ret;
}
void makeFact() {
  fac[0] = ifac[0] = 1;
  for (long long int i = 1; i < 200100; i++) {
    fac[i] = (fac[i - 1] * i) % MOD;
    ifac[i] = powMod(fac[i], MOD - 2);
  }
}
inline void Init() {
  inv[1] = 1;
  for (int i = 2; i < 200100; ++i)
    inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
}
long long int C(long long int a, long long int b) {
  if (a < b) return 0;
  return (((fac[a] * ifac[b]) % MOD) * ifac[a - b]) % MOD;
}
long long int P(long long int a, long long int b) {
  if (a < b) return 0;
  return (fac[a] * ifac[a - b]) % MOD;
}
long long int sum(long long int k) { return (k % MOD) * ((k + 1) % MOD) / 2; }
long long int pascalDivisibility(long long int n, long long int p) {
  long long int ans = 1;
  long long int i = 0;
  while (n) {
    ans = (sum(n % p) % MOD) * (powMod(sum(p), i) % MOD) +
          ((n % p + 1) % MOD) * (ans % MOD);
    ans %= MOD;
    n /= p;
    i++;
  }
  return ans % MOD;
}
bool PointInRectangle(long double x1, long double y1, long double x2,
                      long double y2, long double x, long double y) {
  return (x >= x1 && x <= x2 && y >= y1 && y <= y2);
}
bool InRange(long long int a, long long int b, long long int c) {
  return a >= b && a <= c;
}
string AL = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
long long int arr[200100];
long long int level[200100];
vector<long long int> rang;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, x, mx(0), mn(2 * 1e18);
  cin >> n >> k;
  set<long long int> ss;
  for (int i = 0; i < n; i++)
    cin >> x, arr[x]++, mx = max(mx, x), mn = min(mn, x);
  long long int sum = 0;
  for (int i = 200100 - 1; i > 0; i--) {
    sum += arr[i];
  }
  long long int cnt(0);
  long long int sumi = 0;
  for (long long int i = mx; i > mn; i--) {
    sumi += arr[i];
    level[i] = sumi;
  }
  for (long long int i = mx; i > mn;) {
    long long int sumo = 0;
    while (sumo < k) {
      sumo += level[i];
      i--;
    }
    if (sumo > k) i++;
    cnt++;
  }
  cout << cnt << endl;
  return 0;
}
