You are given a sequence of integers a1,a2,…,an. You need to paint elements in colors, so that:

  • If we consider any color, all elements of this color must be divisible by the minimal element of this color.
  • The number of used colors must be minimized.

For example, it's fine to paint elements [40,10,60]in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.

For example, if a=[6,2,3,4,12]then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10,7,15]then 3 colors are required (we can simply paint each element in an unique color).

Input

The first line contains an integer n (1≤n≤100), where n is the length of the given sequence.

The second line contains n integers a1,a2,…,an (1≤ai≤100). These numbers can contain duplicates.

Output

Print the minimal number of colors to paint all the given numbers in a valid way.

Examples
input
6
10 2 3 5 4 2
output
3
input
4
100 100 100 100
output
1
input
8
7 6 5 4 3 2 2 3
output
4
Note

In the first example, one possible way to paint the elements in 3 colors is:

  • paint in the first color the elements: a1=10 and a4=5,
  • paint in the second color the element a3=3,
  • paint in the third color the elements: a2=2, a5=4 and a6=2.

In the second example, you can use one color to paint all the elements.

In the third example, one possible way to paint the elements in 4 colors is:

  • paint in the first color the elements: a4=4 a6=2 and a7=2,
  • paint in the second color the elements: a2=6, a5=3 and a8=3,
  • paint in the third color the element a3=5,
  • paint in the fourth color the element a1=7.

题意解释:输入n个数,对一个数进行涂色时,被涂色的数的倍数也会被涂色。输出最少涂几次可以涂完所有的数

思路和筛法是一样的,从小的往大的筛,直到筛完为止。

#include <bits/stdc++.h>
using namespace std;
int a[];
int main()
{
int n;
cin>>n;
for(int i=;i<n;++i)
{
int t;
cin>>t;
a[t]=t;
}
int ans=;
for(int i=;i<=;++i)
{
if(a[i])
{
for(int j=i;j<=;j+=i)
{
a[j]=;
}
ans++;
}
}
cout<<ans;
return ;
}

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