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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;
import java.util.function.BiFunction;

public class Main {

    static BiFunction<Long, Long, Long> ADD = (x, y) -> (x + y);

    public static void main(String[] args) throws Exception {
        long startTime = System.nanoTime();
        int t = 1;
        while (t-- > 0) {
            solve();
        }
        long endTime = System.nanoTime();
        err.println("Execution Time : +" + (endTime - startTime) / 1000000 + " ms");
        exit(0);
    }

    static void solve() {
        int n = in.nextInt();
        Map<Integer, Long> counts = new HashMap<>();
        for (int i = 0; i < n; i++) {
            char[] c = in.next().toCharArray();
            Arrays.sort(c);
            counts.merge(Arrays.hashCode(c), 1L, ADD);
        }
        long ans = 0;
        for (long a : counts.values()) {
            ans += a * (a - 1) / 2;
        }
        out.println(ans);
    }

    static void debug(Object... args) {
        for (Object a : args) {
            out.println(a);
        }
    }

    static void y() {
        out.println("YES");
    }

    static void n() {
        out.println("NO");
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public int[] readAllInts(int n) {
            int[] p = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = in.nextInt();
            }
            return p;
        }

        public int[] readAllInts(int n, int s) {
            int[] p = new int[n + s];
            for (int i = s; i < n + s; i++) {
                p[i] = in.nextInt();
            }
            return p;
        }

        public long[] readAllLongs(int n) {
            long[] p = new long[n];
            for (int i = 0; i < n; i++) {
                p[i] = in.nextInt();
            }
            return p;
        }

        public long[] readAllLongs(int n, int s) {
            long[] p = new long[n + s];
            for (int i = s; i < n + s; i++) {
                p[i] = in.nextInt();
            }
            return p;
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }

    static void exit(int a) {
        out.close();
        err.close();
        System.exit(a);
    }

    static InputStream inputStream = System.in;
    static OutputStream outputStream = System.out;
    static OutputStream errStream = System.err;
    static InputReader in = new InputReader(inputStream);
    static PrintWriter out = new PrintWriter(outputStream);
    static PrintWriter err = new PrintWriter(errStream);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

}