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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int sosuu(char a)
{
  if(a=='a')
    return 2;
  if(a=='b')
    return 3;
  if(a=='c')
    return 5;
  if(a=='d')
    return 7;
  if(a=='e')
    return 11;
  if(a=='f')
    return 13;
  if(a=='g')
    return 17;
  if(a=='h')
    return 19;
  if(a=='i')
    return 23;
  if(a=='j')
    return 29;
  if(a=='k')
    return 31;
  if(a=='l')
    return 37;
  if(a=='m')
    return 41;
  if(a=='n')
    return 43;
  if(a=='o')
    return 47;
  if(a=='p')
    return 53;
  if(a=='y')
    return 59;
  if(a=='q')
    return 61;
  if(a=='r')
    return 67;
  if(a=='s')
    return 71;
  if(a=='t')
    return 73;
  if(a=='u')
    return 79;
  if(a=='v')
    return 83;
  if(a=='w')
    return 89;
  if(a=='x')
    return 97;
  if(a=='z')
    return 101;
}

int double_comp(const void* x,const void* y) {
	const double* a=(const double*)x;
	const double* b=(const double*)y;
	if(*a>*b)return 1;
	if(*a<*b)return -1;
	return 0;
}

int main(){
  int i,j;
  int N;
  unsigned long long ans=0,k;
  
  scanf("%d",&N);
  double A[N];
  char tmp[11];
  for(i=0;i<N;i++){
    A[i]=1;
    scanf("%s",tmp);
    for(j=0;j<strlen(tmp);j++){
      A[i]*=sosuu(tmp[j]);
    }
  }
  k=(unsigned long long)1;
  qsort(A,N,sizeof(double),double_comp);
  for(i=0;i<N;i++){
    //printf("%f\n",A[i]);
    if(A[i]==A[i+1]){
      k++;
    }
    else{
      if(k!=(unsigned long long)1)
      ans += k*(k-1)/(unsigned long long)2;
      k=(unsigned long long)1;
    }
  }
  printf("%llu",ans);
  return 0;  
}