Operational countdown Given a non negative integer number \$n\$ output how many steps to reach zero using radicals, divisions or subtractions.
The algorithm

Get digits count ( \$d\$ ) of  \$n\$.

Try the following operations in order:
$$\sqrt[d]{n}$$
$$n/d$$
$$n-d$$

Take the first integer result not equal to  \$n\$.
Floating point errors must be avoided !

Repeat the process with the value obtained until you reach 0.


Example
1500 -&gt; 8
1500 -&gt; 4 digits , ( / ) =&gt; 375 // step 1
375 -&gt; 3 digits , ( / ) =&gt; 125 // step 2
125 -&gt; 3 digits , ( √ ) =&gt; 5 // step 3
5 -&gt; 1 digits , ( - ) =&gt; 4 // step 4
4 -&gt; 1 digits , ( - ) =&gt; 3 // step 5
3 -&gt; 1 digits , ( - ) =&gt; 2 // step 6
2 -&gt; 1 digits , ( - ) =&gt; 1 // step 7
1 -&gt; 1 digits , ( - ) =&gt; 0 // step 8

Input: a non negative integer number. You don't have to handle inputs not supported by your language (obviously, abusing this is a standard loophole)
Output: the number of steps to reach 0
Test cases
n -&gt; steps

0 -&gt; 0
1 -&gt; 1
2 -&gt; 2
4 -&gt; 4
10 -&gt; 6
12 -&gt; 7
16 -&gt; 5
64 -&gt; 9
100 -&gt; 19
128 -&gt; 7
1000 -&gt; 70
1296 -&gt; 7
1500 -&gt; 8
5184 -&gt; 8
10000 -&gt; 133
21550 -&gt; 1000
26720 -&gt; 100
1018080 -&gt; 16
387420489 -&gt; 10

Rules

Input/output can be given by any convenient method.
You can print it to STDOUT, return it as a function result or error message/s.
Either a full program or a function are acceptable.
Standard loopholes are forbidden.
Answers must not fail due to floating point errors.
This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.


Sandbox: https://codegolf.meta.stackexchange.com/a/20518/84844
R, 77 74 73 bytes
-3 bytes thanks to Giuseppe and -1 byte thanks to MarcMush.

f=function(n,d=nchar(n),k=1:n)`if`(d&lt;2,n,1+f(match(n,c(k^d,k*d,k+d))%%n))

Try it online!
Recursive function; avoids floating point issues. The vector k contains all integers from 1 to n. Concatenate k^d, k*d and k+d (yielding a vector of length 3n) and find the position p of the first occurrence of n. Then \$f(n)=1+f(p \mod n)\$. The recursion is initialized by noting that \$f(n)=n\$ for all 1-digit integers (hence the conditioning on d&lt;2).
Works for all the test cases (although 21550 could require you to increase the stack limit on some machines).