Problem:
Score : 200 points

Problem Statement

We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.

In how many ways can we choose three of the sticks with different lengths that can form a triangle?

That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:

L_i, L_j, and L_k are all different.

There exists a triangle whose sides have lengths L_i, L_j, and L_k.

Constraints

1 \leq N \leq 100

1 \leq L_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 \cdots L_N

Output

Print the number of ways to choose three of the sticks with different lengths that can form a triangle.

Sample Input 1

5
4 4 9 7 5

Sample Output 1

5

The following five triples (i, j, k) satisfy the conditions: (1, 3, 4), (1, 4, 5), (2, 3, 4), (2, 4, 5), and (3, 4, 5).

Sample Input 2

6
4 5 4 3 3 5

Sample Output 2

8

We have two sticks for each of the lengths 3, 4, and 5. To satisfy the first condition, we have to choose one from each length.

There is a triangle whose sides have lengths 3, 4, and 5, so we have 2 ^ 3 = 8 triples (i, j, k) that satisfy the conditions.

Sample Input 3

10
9 4 6 1 9 6 10 6 6 8

Sample Output 3

39

Sample Input 4

2
1 1

Sample Output 4

0

No triple (i, j, k) satisfies 1 \leq i < j < k \leq N, so we should print 0.

Solution in Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

program sample
    implicit none
    character(100)::s
    integer(8) :: i,j,m,n,k
    integer(8),allocatable :: x(:),y(:)
  
    read(*,*) n
    allocate(x(n))
    read(*,*)x
    call msort(x)
    m=0
    do i=1,n-2
        do j=i+1,n-1
            do k=j+1,n
                if(x(i)+x(j)>x(k) .and. x(i)/=x(j) .and. x(j)/=x(k))then
                    m=m+1
                end if
            end do
        end do
    end do
    write(*,*)m
  
    stop
    contains
    recursive subroutine msort(x)
    implicit none
    integer(8), intent(inout) :: x(:)
 
    integer(8) :: n, mid, i, j, k
    integer(8), allocatable :: tmp(:)
    
    n = size(x)
    
    if(n == 1) then
      return
    end if
 
    !
    ! 前半と後半に分けてソート
    ! 1~midとmid+1~nをそれぞれソート済みにする
    !
    mid = n / 2
    call msort(x(:mid))
    call msort(x(mid+1:))
 
    !
    ! tmpという配列にxを一時的に保管
    ! マージソートは外部ソート
    !
    allocate(tmp(n))
    tmp = x
 
    ! tmpの後半部分を左右反転
    do k = n, mid+1, -1
      tmp(k) = x(mid + n - k + 1)
    end do
 
    ! iは前半のリストのうちまだ見ていないものの最小値
    ! jは後半のリストのうちまだ見ていないものの最小値
    ! kは見た回数の合計
    i = 1
    j = n
 
    do k = 1, n
      !
      ! 一番左と一番右のうち小さい方をxに入れていく
      ! ここが等号つき不等号なので、安定ソートとなる
      !
      if(tmp(i) <= tmp(j)) then
        x(k) = tmp(i)
        i = i + 1
      else
        x(k) = tmp(j)
        j = j - 1
      end if
    end do
    
    deallocate(tmp)
    return
  end subroutine msort
end program sample