### Input:
Given integers `m` and `n` which are the modular and multiplier for sequence.
### Output:
Return all cyclic sequences for [Digit roots](https://en.wikipedia.org/wiki/Digital_root) of `n` * `i` in base `m`

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


### Example Input:

    9 4

### Example Output:

    1 4 7
    2 5 8

## More details:



    n=4, m=9
    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 [tag:code-golf] rules are applied excepts the answer with smaller sum of digit roots of each individual byte wins.