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.Arrays;
import java.util.Scanner;
public class Main {
	static Scanner scan=new Scanner(System.in);
	static int gcd (int a, int b) {return b>0?gcd(b,a%b):a;}
	static long lcm (int a, int b) {return a*b/gcd(a,b);}
	static int max(int a,int b) {return a>b?a:b;}
	static int min(int a,int b) {return a<b?a:b;}
	static long factorial(int i) {return i==1?1:i*factorial(i-1);}
	static int lower_bound(int a[],int key) {
		int low=0,high=a.length;
		while(low<high) {
			int mid=((high-low)/2)+low;
			if(a[mid]<=key)low=mid+1;
			else high=mid;
		}
		return high;
		}
	static int upper_bound(int a[],int key) {
		int low=0,high=a.length;
		while(low<high) {
			int mid=((high-low)/2)+low;
			if(a[mid]<key)low=mid+1;
			else high=mid;
		}
		return high;
		}
	static boolean isPrime (int n) {
		if (n==2) return true;
		if (n<2 || n%2==0) return false;
		double d = Math.sqrt(n);
		for (int i=3; i<=d; i+=2)if(n%i==0){return false;}
		return true;
	}
	static int upper_division(int a,int b) {
		if(a%b==0) {
			return a/b;
		}
		else {
			return a/b+1;
		}
	}
	static long lupper_division(long a,long b) {
		if(a%b==0) {
			return a/b;
		}
		else {
			return a/b+1;
		}
	}
	static long lmax(long a,long b) {return Math.max(a, b);}
	static long lmin(long a,long b) {return Math.min(a, b);}
	static int[] setArray(int a) {
		int b[]=new int[a];
		for(int i=0;i<a;i++) {
			b[i]=scan.nextInt();
		}
		return b;
	}
		public static void main(String[] args) {
			int a=scan.nextInt();
			String strs[]=new String[a];
			for(int i=0;i<a;i++) {
				String str=scan.next();
				char ch[]=str.toCharArray();
				Arrays.sort(ch);
				strs[i]=String.valueOf(ch);
			}
			Arrays.sort(strs);
			long cnt=0;
			long hoge=1;
			for(int i=1;i<a;i++) {
				String str1=strs[i-1];
				String str2=strs[i];
				if(str1.equals(str2)) {
					hoge++;
				}
				else {
					cnt+=hoge*(hoge-1)/2;
					hoge=1;
				}
			}
			cnt+=hoge*(hoge-1)/2;
			System.out.println(cnt);
		}
}