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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.lang.reflect.Array;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int N = scan.nextInt();
    scan.nextLine();
    long[] result = new long[N];
    byte[][] results = new byte[N][];
    for (int i = 0; i < N; i++) {
      String s = scan.nextLine();
      results[i] = foo(s);
    }

    int count = 0;
    for (int i = 0; i < N - 1; i++) {
      for (int j = i + 1; j < N - 1; j++) {
        if (bar(results[i], results[j + 1])) {
          count++;
        }
      }
    }
    
    System.out.println(count);
  }

  private static byte[] foo(String str) {
    byte[] b = str.getBytes();
    Arrays.sort(b);
    return b;
  }

  private static boolean bar(byte[] b1, byte[] b2) {
    for (int i = 0; i < 10; i++) {
      if (b1[i] != b2[i]) {
        return false;
      }
    }
    return true;
  }

  private static long hoge(String str) {
    long result = 0;
    for (int i = 0; i < str.length(); i++) {
      result += fuga(str.charAt(i));
    }
    return result;
  }

  private static long fuga(char c) {
    switch (c) {
      case 'a':
        return 1;
      case 'b':
        return 2;
      case 'c':
        return 4;
      case 'd':
        return 8;
      case 'e':
        return 16;
      case 'f':
        return 32;
      case 'g':
        return 64;
      case 'h':
        return 128;
      case 'i':
        return 256;
      case 'j':
        return 512;
      case 'k':
        return 1024;
      case 'l':
        return 2048;
      case 'm':
        return 4096;
      case 'n':
        return 8192;
      case 'o':
        return 16384;
      case 'p':
        return 32768;
      case 'q':
        return 65536;
      case 'r':
        return 131072;
      case 's':
        return 262144;
      case 't':
        return 524288;
      case 'u':
        return 1048576;
      case 'v':
        return 2097152;
      case 'w':
        return 4194304;
      case 'x':
        return 8388608;
      case 'y':
        return 16777216;
      case 'z':
        return 33554432;
      default:
        throw new RuntimeException();
    }
  }

}