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: Accepted)

import java.util.*;
import java.io.*;
 
public class Main {
	private static void solve(){
		//Implement solution here.
		int n = ni();		
		Map<String,Long> m = new HashMap<>();
		long result = 0L;
		char[] tmp = new char[10];		
		for(int i = 0; i < n; i++) {
			tmp = next().toCharArray();
			Arrays.sort(tmp);
			String tmps = new String(tmp);
			if(m.containsKey(tmps)) {
				m.put(tmps, m.get(tmps) + 1L);
			}else {
				m.put(tmps, 1L);
			}
		}
		
		for(String key:m.keySet()) {
			long v = m.get(key);
			if(v > 1) {
				result += v * (v - 1) / 2;
			} 
		}
		
		System.out.println(result);
	}

 
	//Switch input source (stdin/file)
	private static BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
	public static void main(String[] args){
		String debugDataPath = System.getenv("DD_PATH");        
		if(debugDataPath != null){
			try{
				br = new BufferedReader(new InputStreamReader(new FileInputStream(new File(debugDataPath))));
			}catch(Exception e){
				throw new RuntimeException(e);
			}
		} 
		solve();
	}
 
	//Input read utility
	private static StringTokenizer tokenizer = null;
	public static String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
	//Get next single int
	private static int ni() {
		return Integer.parseInt(next());
	}
	//Get next single long
	private static long nl() {
		return Long.parseLong(next());
	}
	//Get next single double
	private static double nd() {
		return Double.parseDouble(next());
	}
	//Get next int array from one line
	private static int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}
	//Get next char array from one line
	private static char[] ns() {
		return next().toCharArray();
	}
	//Get next long array from one line
	private static long[] nal(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}
}