CodeForces 975A Aramic script

Description：

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 $n$ ($1\le n\le {10}^3$) — 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

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".

題目大意：

求給的一系列字符串中有多少個“根”， 其實就是有多少種不同類型的字符串。所謂同一種類型的字符串比如 a aa aaa 它們都屬於‘a’， ab abb 都屬於‘ab’。 根的字母必須只出現一次， 根的全排列也同屬一種。

解題思路：

當時這題寫的真的很辣雞。

賽後在這裏先感慨一下set的好用！ 然後題就寫出來了。。。 把字符串一個個塞進set裏， 判斷是否出現過相同的根， 如果沒出現放進root裏， 最後統計個數即可。

代碼：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <utility>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int dir[9][2] = { 0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1, 0, 0 };
const int inf = 0x3f3f3f;
const ll ll_inf = (ll) 1e15;
const double pi = acos(-1.0);
const int mod = (ll) 1e9 + 7;
const int Max = (int) 1e3 + 51;
const ld eps = 1e-10;

/*tools:
*ios::sync_with_stdio(false);
*freopen("input.txt", "r", stdin);
*/

set < char> S, root[Max];

int main() {
    int n;
    int cnt = 0;
    string st;

    cin >> n;
    for (int i = 0; i < n; ++i) {
        S.clear();

        cin >> st;
        for (int j = 0; j < st.size(); ++j)
            S.insert(st[j]);
        set<char>::iterator l;
        int f = 0;
        for (int k = 0; k < cnt; ++k) {
            if (root[k] == S) {
                f = 1;
                break;
            }
        }
        if (!f) {
            root[cnt] = S;
            cnt++;
        }
    }
    printf("%d\n", cnt);
    return 0;
}