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).
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
from math import gcd
import io
import os
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect

#?############################################################


def isPrime(x):
    for i in range(2, x):
        if i*i > x:
            break
        if (x % i == 0):
            return False
    return True

#?############################################################


def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den, p - 2, p)) % p


#?############################################################

def primeFactors(n):
    l = []
    while n % 2 == 0:
        l.append(2)
        n = n / 2
    for i in range(3, int(math.sqrt(n))+1, 2):
        while n % i == 0:
            l.append(int(i))
            n = n / i
    if n > 2:
        l.append(n)
    return list(set(l))


#?############################################################

def power(x, y, p):
    res = 1
    x = x % p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res

#?############################################################


def sieve(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime


#?############################################################

def digits(n):
    c = 0
    while (n > 0):
        n //= 10
        c += 1
    return c

#?############################################################


def ceil(n, x):
    if (n % x == 0):
        return n//x
    return n//x+1

#?############################################################


def mapin():
    return map(int, input().split())

#?############################################################


input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
mm =200009
n, k = map(int, input().split())
l = [int(x) for x in input().split()]
ll = [0]*(mm)
for i in l:
    ll[i]+=1




ans = 0
mmm = min(l)
cap = 0
i= mm-1
h = max(l)
ss = 0
while(True):
    x = ss-cap*(i-1)
    if(x> k):
        ans+=1
        h= i
        ss = i*cap
     
    i-=1   
    if(i == mmm):
        break
    cap += ll[i]
    ss+= ll[i]*i


if(i!=h):
    ans+=1

            
print(ans)