Aramic script

In Aramic language words can only represent objects.

Words in Aramic have special properties:

• A word is a root if it does not contain the same letter more than once.
• A root and all its permutations represent the same object.
• The root $\mathrm{xx of\; a\; word yy is\; the\; word\; that\; contains\; all\; letters\; that\; appear\; in yy in\; a\; way\; that\; each\; letter\; appears\; once.\; For\; example,\; the rootof\; "aaaa",\; "aa",\; "aaa"\; is\; "a",\; the root of\; "aabb",\; "bab",\; "baabb",\; "ab"\; is\; "ab".}$
• Any word in Aramic represents the same object as its root.

You have an ancient script in Aramic. What is the number of different objects mentioned in the script?

Input

The first line contains one integer $\mathrm{nn (1\le n\le 1031\le n\le 103) —\; the\; number\; of\; words\; in\; the\; script.}$

The second line contains $\mathrm{nn words s1,s2,\dots ,sns1,s2,\dots ,sn —\; the\; script\; itself.\; The\; length\; of\; each\; string\; does\; not\; exceed 103103.}$

It is guaranteed that all characters of the strings are small latin letters.

Output

Output one integer — the number of different objects mentioned in the given ancient Aramic script.

Examples

input

Copy

5
a aa aaa ab abb

output

Copy

2

input

Copy

3
amer arem mrea

output

Copy

1

Note

In the first test, there are two objects mentioned. The roots that represent them are "a","ab".

In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".

意思是问有几个root,root的意思是一个字符串中所有的不重复字母，无关顺序。 比如abbbcd的root是abcd。

面向STL编程，写的具蠢，用了一个map标记出现过的字符，然后扫了一遍，最后用set维护。

#include <iostream>
#include <bits/stdc++.h>
#define maxn 300005
using namespace std;

string findroot(string a)
{   int i;string b;
    map<char,bool> mp;
    for(i=0;i<a.size();i++)
    {
        mp[a[i]]=1;
    }
    for(i='a';i<='z';i++)
    {
        if(mp[i])
        {
           b.push_back(i);
        }
    }
    return b;
}
string a[1005];
int main()
{
    int n,i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    int j;
    bool vis[1005]={0};
    set<string> s;
    for(i=1;i<=n;i++)
    {
        s.insert(findroot(a[i]));
    }
        int cnt=s.size();
    cout<<cnt<<endl;
    return 0;
}