Problem:
Score : 400 points

Problem Statement

Given is a number sequence A of length N.

Find the number of integers i \left(1 \leq i \leq N\right) with the following property:

For every integer j \left(1 \leq j \leq N\right) such that i \neq j , A_j does not divide A_i.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^6

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N

Output

Print the answer.

Sample Input 1

5
24 11 8 3 16

Sample Output 1

3

The integers with the property are 2, 3, and 4.

Sample Input 2

4
5 5 5 5

Sample Output 2

0

Note that there can be multiple equal numbers.

Sample Input 3

10
33 18 45 28 8 19 89 86 2 4

Sample Output 3

5

Solution in Lua (LuaJIT 2.1.0) (judge verdict: Accepted)

-- RayStark氏のコードを参考にした

local read = io.read
local floor = math.floor
local N = read("*n")
local A = {}
for i = 1, N do
	A[i] = read("*n")
end
table.sort(A)

local A_MAX = A[N]

local cant_div_s = {}
for i = 1, N do
	-- なんでcant_divs[i],(1<=i<=N)でないか
	-- A[i]にない数字は考えなくて良いから
	cant_div_s[A[i]] = true
end

local last_value = 0
for i = 1, N do
	-- cant_divs全部回したら時間かかるんじゃないのか？
	-- []のなかにA[i],(1<=i<=N)を使ってるのでN回の計算で済む
	if cant_div_s[A[i]] and last_value == A[i] then
		cant_div_s[A[i]] = false
	elseif cant_div_s[A[i]] then
		-- なんでj=1でなく2か
		-- A_MAXまでのA[i]より大きい倍数は倍数だから
		-- TODO: ここがボトルネックになっているためここの最適化
		for j = 2, floor(A_MAX / A[i]) do
			cant_div_s[A[i] * j] = false
		end
		-- なんで最後に処理した数字を残しておく必要があるか
		-- 同じ数が存在する場合にはじくため
		last_value = A[i]
	end
end

local num = 0
for i = 1, N do
	if cant_div_s[A[i]] then
		num = num + 1
	end
end

print(num)