CF1209A Paint the Numbers

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[105];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;++i)
    {
        int t;
        cin>>t;
        a[t]=t;
    }
    int ans=0;
    for(int i=1;i<=100;++i)
    {
        if(a[i])
        {
            for(int j=i;j<=100;j+=i)
            {
                a[j]=0;
            }
            ans++;
        }
    }
    cout<<ans;
    return 0;
}