题目链接：

http://codeforces.com/contest/975/problem/A

题目：

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 $n(1\leq n \leq 10^3)$ — the number of words in the script.

The second line contains $n$ words $s_1,s_2,\dots,s_n$ — the script itself. The length of each string does not exceed $10^3$.

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

5
a aa aaa ab abb

output

2

input

3
amer arem mrea

output

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".

题意：

  一个单词的“根”为对这个单词所包含的字符排序去重之后剩下的内容，如ccab的“根”为abc。然后给你n个单词，问这些单词里有多少个不同的根。

思路：

  读入每个字符串之后sort一下然后unique一下加到set里即可。

实现：

#include <bits/stdc++.h>
int n;
std::set<std::string> set;
char str[1007];
int main() {
//    freopen("in.txt", "r", stdin);
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf(" %s", str);
        int len = int(strlen(str));
        std::sort(str, str + len);
        len = int(std::unique(str, str + len) - str);
        str[len] = '\0';
        set.insert(str);
    }
    printf("%d\n", int(set.size()));
    return 0;
}