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 (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

program prob4
    implicit none
    integer::i, j, l, n, tmp
    integer(8)::ans
    integer, dimension(10)::c
    integer, allocatable::p(:,:)
    character(len=10)::S
    read(*,*) n
    allocate(p(n,10))
    do i = 1, n
        read(*,*) S
        c = 0
        do j = 1,10
            c(j) = ichar(S(j:j)) - ichar('a') + 1
        end do
        call heapsort(10,c)
        p(i,:) = c
    end do
    call lsort(n,p)
    ans = 0
    tmp = 0
    do i = 2,n
        do j = 1, 10
            if(p(i,j) .ne. p(i-1,j)) then
                tmp = 0
                exit
            else if(j == 10) then
                tmp = tmp + 1
                ans = ans + tmp
            end if
        end do
    end do
    write(*,*) ans
    stop
contains
    subroutine heapsort(n,array)
        implicit none
        integer,intent(in) :: n
        integer,intent(inout) :: array(1:n)
    
        integer ::i,k,j,l
        double precision :: 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

    subroutine lsort(n,array)
        implicit none
        integer,intent(in) :: n
        integer,intent(inout) :: array(n,10)
    
        integer ::i,k,j,l,m,mode
        integer::t(10)
    
        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
                    do m = 1, 10
                        if(array(j,m).lt.array(j+1,m))j=j+1
                        exit
                    end do
                endif
                mode = 0
                do m = 1, 10
                    if(t(m).lt.array(j,m)) then
                        mode = 1
                        exit
                    end if
                end do
                if (mode==1)then
                    array(i,:)=array(j,:)
                    i=j
                    j=j+j
                else
                    j=k+1
                endif
            enddo
            array(i,:)=t
        enddo
    
        return
    end subroutine lsort
end program