Digit sum without loops

You are given a non-negative integer $n$. Find the sum of its digits without loops, using recursion.

A single integer $n$ ($0 \le n \le 10^{18}$).

A single integer — the sum of the digits of $n$.

The last digit is $n \bmod 10$, the remaining digits are $n \operatorname{div} 10$. So $S(n) = n \bmod 10 + S(n \operatorname{div} 10)$, with the base case $S(n) = n$ for $n < 10$. For example, $S(12345) = 5 + S(1234) = \ldots = 15$. $10^{18}$ fits in a 64-bit type.