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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

namespace cs
{
    public class Program
    {
        public static void Main(string[] args)
        {
            /* 
            var N = long.Parse(Console.ReadLine());
            var cities = Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);
            var brave = Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);
            */

            //Console.WriteLine(outer);
            //SLV(new long[]{13, 3});
            SLV(Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse));
            //SLV(Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse));
            //SLV(Console.ReadLine());
            Console.ReadKey();
        }

        public static void SLV(long[] input){

            var alph = "abcdefghijklmnopqrstuvwxyz".ToCharArray();

            var N = input[0];
            var x = new Dictionary<String, long>();
            char[] str;
            string str1;

            for(int i = 0; i < N; i++){

                str = Console.ReadLine().ToCharArray();
                Array.Sort(str);
                str1 = new string(str);
                if(x.ContainsKey(str1)) x[str1] += x[str1]+1;
                else x[str1] = 0;
            }

            Console.WriteLine(x.Values.Sum());


            

            return;
        }




        public static long modder(){
            return (long)Math.Pow(10,9) + 7;
        }

        public static long GCD(long n, long m){
            
            if(m % n == 0) return n;
            else return GCD(m%n, n);

        }

        public static long LCD(long n, long m){

            return m * n / GCD(n,m);
        }

        
        // いやなんか違うけどねこれ
        public static long lower_bound(long[] array, long input){
            var left = 0;
            var right = array.Length-1;
            var mid =0;
            while(right >= left){
                mid = left + (right - left)/2;
                if(array[mid] == input){
                    return mid;
                }
                if(array[mid] > input){
                    right = mid-1;
                }else{
                    left = mid+1;
                }
            }
    
            return right;
        }

        
        class UfTree{
            public long[] parents{get;set;}
            public long[] size{get;set;} //sizeはrootのみが最新のものを保持する
    
            public UfTree(long n){
                parents = new long[n];
                size = new long[n];
                for(long i = 0; i < n; i++){
                    parents[i] = i;
                    size[i] = 1;
                } 
            }
    
            public long getRoot(long x){
                if(parents[x] == x){
                    return x;
                }else{
                    parents[x] = getRoot(parents[x]);
                    return parents[x];
                }
            }
    
            public long unite(long x, long y){
                var rx = getRoot(x);
                var ry = getRoot(y);
                if(rx == ry){
                    return rx;
                } else{
                    var sx = getSize(rx);
                    var sy = getSize(ry);
                    if(sx > sy){
                        size[rx] = sx+sy;
                        parents[ry] = rx;
                        return rx;
                    }else{
                        size[ry] = sx+sy;
                        parents[rx] = ry;
                        return ry;
                    }
                }
                
            }
    
            public long getSize(long x){
                return size[getRoot(x)];
            }
    
            
    
            public bool isSameTree(long x, long y){
                return getRoot(x) == getRoot(y);
            }
 
        }
    }


    
}