A > B > C > A

Given 2 dice $A,B$, we define $\mathfrak{P}(A,B)$ as the probability of rolling a strictly higher value with $A$ than $B$. If $A=\{2,2,4,4,9,9\}$ and $B=\{1,1,6,6,8,8\}$ we have $\mathfrak{P}(A,B)=\frac{5}{9}>\frac{1}{2}$

Thus the probability to win with $A$ is greater than with $B$.
We write: $A>B$

We add a third die $C=\{3,3,5,5,7,7\}$.
We have:
$\mathfrak{P}(B,C)=\frac{5}{9}$
$\mathfrak{P}(C,A)=\frac{5}{9}$

Therefore: $A>B>C>A$

A triple of dice is called non-transitive if:
$A>B>C>A$
$\mathfrak{P}(A,B)=\mathfrak{P}(B,C)=\mathfrak{P}(C,A)$

Of course if $(A,B,C)$ is such a triple, so is $(B,C,A)$ and $(C,A,B)$.
These 3 triples are considered equivalent.

We consider dice with values in range $\{1,2,3,4,5,6,7,8\}$.

How many dictinct non-transitive triples are there?
There are triples where all faces for $A$ have a distinct value. What is the last one in lexicographic order?
Between the 3 possible representation for a triple $(A,B,C),(B,C,A)\text{ and }(C,A,B)$, we choose the one where $A$ is the first in lexicographic order

Answer format: Count/(die1)(die2)(die3)
Example : 9999/(9,9,9,9,9,9)(9,9,9,9,9,9)(9,9,9,9,9,9)

You are given there are 104 distinct non-transitive triples for the set $\{1,2,3,4,5,6\}$

[My timing: 80 sec]