Input:

Given integers m and n which are the modular and multiplier for sequence.

Output:

Return all cyclic sequences of length greater than one for Digit roots of n * i in base m + 1

• \$1\$ ≤ \$i\$ ≤ \$m\$
• \$1\$ ≤ \$DR(n*i) \$ ≤ \$m\$
• \$DR(i) = i \$
• \$DR(-m) = 0 \$
• \$|DR(-x)| \equiv DR(x) \equiv -DR(x)\$ (mod \$ m) \$
• \$DR(a+b) = DR(DR(a)+DR(b))\$
• \$DR(a-b) \cong (DR(a)-DR(b))\$ (mod \$ m) \$

Example Input:

9 4

Example Output:

1 4 7
2 5 8

More details:

m=9, n=4
DR(4*1) -> 4
DR(4*4) -> 7
DR(4*7) -> 1 = first i

A cycle happens when going through numbers 1 trough m taking digit root of n * i and then digit root of n * result of previous call and so on until returns to the first i.

Note that if there is no cycle taking digit root of n * i would simply result to i.

So we store this sequence in something like a hash set for all the i's and then return all the sequences.

Challenge

All the normal code-golf rules are applied excepts the answer with smaller sum of digit roots of each individual byte wins.

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