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).
# -*- coding: utf-8 -*-

import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys

"""
created by shhuan at 2018/10/19 23:16

"""

def canSliceToHeight(highest, targetHeight, start, heights, maxSliceCount):
    total = (highest - targetHeight) * start
    if total > maxSliceCount:
        return False

    for i in range(start, len(heights)):
        h = heights[i]
        if h <= targetHeight:
            break
        total += heights[i] - targetHeight

    return total <= maxSliceCount

def findNextHighest(currentHighest, start, heights, maxSliceCount):
    lo, hi = 0, currentHighest

    while lo < hi:
        m = (lo + hi) // 2
        if canSliceToHeight(currentHighest, m, start, heights, maxSliceCount):
            hi = m
        else:
            lo = m + 1


    return lo

def solve(N, K, H):
    H.sort(reverse=True)
    target = H[-1]
    steps = 0

    start = 1
    highest = H[0]
    done = False
    while highest > target and not done:
        nextHighest = findNextHighest(highest, start, H, K)

        if H[start] < nextHighest:
            slicedHeightPerStep = highest - nextHighest
            localSteps = (highest - H[start]) // slicedHeightPerStep
            steps += localSteps
            highest -= slicedHeightPerStep * localSteps
        else:
            nextStart = start
            for i in range(start, N):
                if H[i] < nextHighest:
                    nextStart = i
                    break

            start = nextStart
            steps += 1
            highest = nextHighest


    return steps


N, K = map(int, input().split())
H = [int(x) for x in input().split()]


print(solve(N, K, H))





