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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections;
using static System.Console;
using static System.Math;
 
public class Hello{
    public static void Main(){
        string alpha = "abcedfghijklmnopqrstuvwxyz";
        int num = rint();
        string[,] list = new string[130,100010];
        int[,] listnum = new int[130,100010];
        long total = 0;
        for(int i=0;i<num;i++){
            string read = rstr();
            char[] chan = new char[10];
            for(int j=0;j<10;j++){
                chan[j] = read[j];
            }
            Array.Sort(chan);
            string change = "";
            for(int j=0;j<10;j++){
                change += chan[j];
            }
            int fir = alpha.IndexOf(chan[0]) +alpha.IndexOf(chan[1]) +alpha.IndexOf(chan[5])+ alpha.IndexOf(chan[8])+ alpha.IndexOf(chan[9]);
            for(int j=0;j<100010;j++){
                if(list[fir,j] == null){
                    list[fir,j] = change;
                    total += listnum[fir,j];
                    listnum[fir,j]++;
                    break;
                }else if(list[fir,j] == change){
                    total += listnum[fir,j];
                    listnum[fir,j]++;
                    break;
                }
            }
        }
        WriteLine(total);
    
    
    
    
    }
    //各種読取
    static string rstr(){
        return ReadLine();
    }
    static int rint(){
        return int.Parse(ReadLine());
    }
    static long rlong(){
        return long.Parse(ReadLine());
    }
    static string[] stra(){
        return ReadLine().Split(' ');
    }
    static int[] inta(){
        string[] read_str_array = ReadLine().Split(' ');
        int[] return_int_array = new int[read_str_array.Length];
        for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_int_array[countup_i] = int.Parse(read_str_array[countup_i]);
        }
        return return_int_array;
    }
    static int[,] inta2(int num_array,int in_array){
        int[,] int_array2 = new int[num_array,in_array];
        for(int i=0;i<num_array;i++){
            int[] temp_array = inta();
            for(int j=0;j<in_array;j++){
                int_array2[i,j] = temp_array[j];
            }
        }
        return int_array2;
    }
    static long[] longa(){
        string[] read_str_array = ReadLine().Split(' ');
        long[] return_long_array = new long[read_str_array.Length];
        for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_long_array[countup_i] = long.Parse(read_str_array[countup_i]);
        }
        return return_long_array;
    }
    static double[] doublea(){
        string[] read_str_array = ReadLine().Split(' ');
        double[] return_double_array = new double[read_str_array.Length];
        for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_double_array[countup_i] = long.Parse(read_str_array[countup_i]);
        }
        return return_double_array;
    }
    // A/B小数点以下切上
    static int divideup(int divided,int divid){
        int temp_divide = divided/divid;
        if(divided % divid > 0){
            temp_divide++;
        }
        return temp_divide;
    }
}