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;
using System.Collections.Generic;
using System.Collections.Specialized;
using System.Linq;
using System.Numerics;

namespace AtCoderWorkspace
{
    class Program
    {
        static void Main(string[] args)
        {
            var cin = new Scanner();
            var n = cin.nextInt();

            var dict = new Dictionary<string, int>();
            var ans = 0;
            for (int i = 0; i < n; i++)
            {
                var s = cin.next().ToCharArray();
                Array.Sort(s);
                var ss = new string(s);

                if (dict.ContainsKey(ss))
                {
                    dict[ss]++;
                    ans += dict[ss];
                }
                else
                {
                    dict.Add(ss, 0);
                }
            }

            Console.WriteLine(ans);
        }
    }

    static class C
    {
        public static long LowerBound(long[] a, long key)
        {
            long l = 0;
            long r = a.Length;
            long mid;
            while (l < r)
            {
                mid = ((r - l) / 2) + l;
                if (a[mid] <= key)
                    l = mid + 1;
                else
                    r = mid;
            }
            return l;
        }
    }

    static class Subsequence
    {
       /// <summary>
       /// 最長増加部分列
       /// </summary>
       /// <param name="a">long[]</param>
       /// <returns></returns>
        public static long LIS(long[] a)
        {
            var dp = Enumerable.Range(0, a.Length).Select(x => long.MaxValue).ToArray();
            for (int i = 0; i < a.Length; i++)
            {
                dp[C.LowerBound(dp, a[i])] = a[i];
            }
            return C.LowerBound(dp, long.MaxValue-1);
        }
    }

    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }

        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            if (s.Length == 0) return next();
            i = 0;
            return s[i++];
        }

        public int nextInt()
        {
            return int.Parse(next());
        }
        public int[] ArrayInt(int N, int add = 0)
        {
            int[] Array = new int[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextInt() + add;
            }
            return Array;
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public long[] ArrayLong(int N, long add = 0)
        {
            long[] Array = new long[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextLong() + add;
            }
            return Array;
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }


        public double[] ArrayDouble(int N, double add = 0)
        {
            double[] Array = new double[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextDouble() + add;
            }
            return Array;
        }
    }
}