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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program ccc

implicit none
integer :: n, i, j
integer :: co, res1 = 0, res2 = 0
integer,dimension(26) :: alp = [2,3,5,7,11,13,17,19,&
&23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101]
integer,allocatable,dimension(:) :: bet
character(10),allocatable,dimension(:) :: s

read*, n
allocate(s(n))
allocate(bet(n))
bet = 1

read*, s

do i=1, n
 do j=1,10
 bet(i)=bet(i)*(alp(ichar(s(i)(j:j))-96))
 end do
end do

call heapsort(n,bet)

co=bet(1)
do i=2,n
 if(co==bet(i)) then
 res1=res1+1
 else
 co=bet(i)
 res2=res2+res1*(res1+1)/2
 res1=0
 end if
end do

res2=res2+res1*(res1+1)/2

write(*,'(i0)') res2

end program


subroutine heapsort(n,array)
  implicit none
  integer,intent(in) :: n
  integer,intent(inout) :: array(1:n)
 
  integer ::i,k,j,l
  integer :: t
 
  if(n.le.0)then
     write(6,*)"Error, at heapsort"; stop
  endif
  if(n.eq.1)return

  l=n/2+1
  k=n
  do while(k.ne.1)
     if(l.gt.1)then
        l=l-1
        t=array(L)
     else
        t=array(k)
        array(k)=array(1)
        k=k-1
        if(k.eq.1) then
           array(1)=t
           exit
        endif
     endif
     i=l
     j=l+l
     do while(j.le.k)
        if(j.lt.k)then
           if(array(j).lt.array(j+1))j=j+1
        endif
        if (t.lt.array(j))then
           array(i)=array(j)
           i=j
           j=j+j
        else
           j=k+1
        endif
     enddo
     array(i)=t
  enddo

  return
end subroutine heapsort