Problem:
Score : 300 points

Problem Statement

We will call a string obtained by arranging the characters contained in a string a in some order, an anagram of a.

For example, greenbin is an anagram of beginner. As seen here, when the same character occurs multiple times, that character must be used that number of times.

Given are N strings s_1, s_2, \ldots, s_N. Each of these strings has a length of 10 and consists of lowercase English characters. Additionally, all of these strings are distinct. Find the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Constraints

2 \leq N \leq 10^5

s_i is a string of length 10.

Each character in s_i is a lowercase English letter.

s_1, s_2, \ldots, s_N are all distinct.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N

Output

Print the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Sample Input 1

3
acornistnt
peanutbomb
constraint

Sample Output 1

1

s_1 = acornistnt is an anagram of s_3 = constraint. There are no other pairs i, j such that s_i is an anagram of s_j, so the answer is 1.

Sample Input 2

2
oneplustwo
ninemodsix

Sample Output 2

0

If there is no pair i, j such that s_i is an anagram of s_j, print 0.

Sample Input 3

5
abaaaaaaaa
oneplustwo
aaaaaaaaba
twoplusone
aaaabaaaaa

Sample Output 3

4

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>


/***  それぞれ長さが１０で英小文字からなるN個の文字列s1, s2,..., snが与えられ、  ***\
      これらの文字列がすべて異なる。このとき二整数i, j(1 <= i < j <= N)に対し  
\***  てsiがsjのアナグラムであるようなものの個数を求めて表示するプログラム        ***/


unsigned long int anagram(unsigned long int n, char **s) {
  unsigned long int i, j, k;                     // カウンタ用
  unsigned long int flag;
  unsigned long int num = 0;                   // アナグラムであるようなものの個数
  unsigned long int alph[26];              // alph[0]-->a, alph[1]-->b,..., alph[25]-->zのようにアルファベットに対応する。alph[i]はaからi番目の英小文字の個数を表す。（初期値は０）

  for(i = 0; i < n-1; i++) {
    // ----------------------------------------------------------------------------------
    for(j = i+1; j < n; j++) {
      flag = 0;
      for(k = 0; k < 26; k++)
	  alph[k] = 0;             // ０に初期化
      for(k = 0; k < 10; k++) { // ある文字列に含まれるアルファベットのそれぞれの数を求める
	alph[(unsigned long int)s[i][k] - 97]++;   // s[i][k]に対応するアルファベットの数をインクリメント
	//printf("%c, s[%d][%d]is%d\n", s[i][k], i, k, (int)s[i][k]);
      }
      for(k = 0; k < 10; k++)
	alph[(unsigned long int)s[j][k] - 97]--;
      
      for(k = 0; k < 26; k++) {
	if(alph[k] != 0) {
	  flag = 0;
	  break;
	} else {
	  flag = 1;
	}
      }
      if(flag == 1)
	num += 1;
    }
    // -----------------------------------------------------------------------------------
  }

  return num;
} 

int main(void) {
  unsigned long int i, j;                         // カウンタ用
  unsigned long int n;                         // 入力される文字列の個数
  char buf[256];                 // 入力用配列
  unsigned long int num;                       // s_iがs_jのアナグラムであるようなものの個数（1 <= i < j <= n)

  fgets(buf, sizeof(buf), stdin);
  n = atoi(buf);

  char **s;
  s = malloc(sizeof(char *) * n);// n行分の領域を確保
  for(i = 0; i < n; i++)
    s[i] = malloc(sizeof(char) * 10); // 1行分の領域を確保（１０文字分＋NULL文字分）
  
  for(i = 0; i < n; i++) {       // n行分の文字を読み込む
    fgets(buf, sizeof(buf), stdin);
    for(j = 0; j < 10; j++)
      s[i][j] = buf[j];
    //printf("s[%d] --> %s\n", i, s[i]);
  }
  
  num = anagram(n, s);
  printf("%lu\n", num);
  
  return 0;
}