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{
	public static void main(String sp[])throws IOException{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		ArrayList<String> al = new ArrayList<>();
		for(int i=0;i<n;i++){
			String temp = sc.next();
			char arr[] = temp.toCharArray();
			Arrays.sort(arr);
			String st = new String(arr);
			al.add(st);
		}
		Collections.sort(al);
		long count=0;
		long total=0;
		for(int i=0;i<n-1;i++){
			if(al.get(i).equals(al.get(i+1))){
				total++;
			}else{
				total+=1;
				count+=((total*(total-1))/2);
				total=0;
			}
		}
		total++;
		count+=((total*(total-1))/2);
		System.out.println(count);
		
	}
	
	
	static class Reader
		{
			final private int BUFFER_SIZE = 1 << 16;
			private DataInputStream din;
			private byte[] buffer;
			private int bufferPointer, bytesRead;
			BufferedReader br;
			public Reader()
			{
				din = new DataInputStream(System.in);
				buffer = new byte[BUFFER_SIZE];
				bufferPointer = bytesRead = 0;
				br = new BufferedReader(new InputStreamReader(System.in));
			}
	 
			public Reader(String file_name) throws IOException
			{
				din = new DataInputStream(new FileInputStream(file_name));
				buffer = new byte[BUFFER_SIZE];
				bufferPointer = bytesRead = 0;
			}
	 
			public String readLine() throws IOException
			{
				byte[] buf = new byte[64]; // line length
				int cnt = 0, c;
				while ((c = read()) != -1)
				{
					if (c == '\n')
						break;
					buf[cnt++] = (byte) c;
				}
				return new String(buf, 0, cnt);
			}
			
			
			
	 
			public int nextInt() throws IOException
			{
				int ret = 0;
				byte c = read();
				while (c <= ' ')
					c = read();
				boolean neg = (c == '-');
				if (neg)
					c = read();
				do
				{
					ret = ret * 10 + c - '0';
				} while ((c = read()) >= '0' && c <= '9');
	 
				if (neg)
					return -ret;
				return ret;
			}
	 
			public long nextLong() throws IOException
			{
				long ret = 0;
				byte c = read();
				while (c <= ' ')
					c = read();
				boolean neg = (c == '-');
				if (neg)
					c = read();
				do
				{
					ret = ret * 10 + c - '0';
				}
				while ((c = read()) >= '0' && c <= '9');
				if (neg)
					return -ret;
				return ret;
			}
	 
			public double nextDouble() throws IOException
			{
				double ret = 0, div = 1;
				byte c = read();
				while (c <= ' ')
					c = read();
				boolean neg = (c == '-');
				if (neg)
					c = read();
	 
				do
				{
					ret = ret * 10 + c - '0';
				}
				while ((c = read()) >= '0' && c <= '9');
	 
				if (c == '.')
				{
					while ((c = read()) >= '0' && c <= '9')
					{
						ret += (c - '0') / (div *= 10);
					}
				}
	 
				if (neg)
					return -ret;
				return ret;
			}
	 
			private void fillBuffer() throws IOException
			{
				bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
				if (bytesRead == -1)
					buffer[0] = -1;
			}
	 
			private byte read() throws IOException
			{
				if (bufferPointer == bytesRead)
					fillBuffer();
				return buffer[bufferPointer++];
			}
			StringTokenizer st;
			String next() {
				while (st == null || !st.hasMoreTokens()) {
					try {
						st = new StringTokenizer(br.readLine());
					} catch (Exception r) {
						r.printStackTrace();
					}
				}
				return st.nextToken();
			}
	 
	 
			public void close() throws IOException
			{
				if (din == null)
					return;
				din.close();
			}
		}
}