Codeforces Round #478 (Div. 2) A. Aramic script

A. Aramic script

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 x

of a word y is the word that contains all letters that appear in y

• in a way that each letter appears once. For example, the root of "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 n

(1≤n≤103

) — the number of words in the script.

The second line contains n

words s1,s2,…,sn — the script itself. The length of each string does not exceed 103

.

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".

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<string>
#include<cmath>
using namespace std;
typedef long long ll;
char str[1005];
set<char> se;
map<string,int> ma;
int main()
{
	int n;
	
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%s",str);
		int len=strlen(str);
		for(int j=0;j<len;j++){
			se.insert(str[j]);
		}
		string s;
		set<char>::iterator ite1 = se.begin();
		s=*ite1;
		ite1++;
		for(;ite1!=se.end();ite1++)
    	{
        	s+=*ite1;
    	}
    	ma[s]++;
		se.clear();
	}
	printf("%d\n",ma.size());
	
	return 0;