Regular expression for $\{x : \#_{01}(x) \mod 3 = 0\}$ Let $L_1$ be the language over alphabet $\{0, 1\}$ defined by $L_1 = \{x : \#_{01}(x) \mod 3 = 0\}$.
Give a regular expression that denotes $L_1$, and justify its correctness

Attempt
Believe this is asking for number of $01$'s in $x$ but $|x|$ must be an even number. Like $x = 0101$, $|x| = |0101| = 4 \mod 3 = 1$ wouldn't work. However $x = 010101, |x| = 6$ would work?
How would I make a regex for this?
$R_1 = (01)^*$. How would I do the mod condition?
Note that any amount of ones at the start or zeroes at the end has no influence on the number of $01$ pairs. Then in the middle you want any multiple of three of $01$ pairs, giving:

$$1^*(0^+1^+0^+1^+0^+1^+)^*0^*$$