13. Given three different prime numbers $p$, $q$, and $r$ that satisfy $p q r = 18 \underbrace{99 \ldots 9}_{n \uparrow 9} 62$, define $f(n)$ as the sum of the digits of the natural number $n$. Then $f(p) + f(q) + f(r) - f(p q r) =$ $\qquad$
【Analysis】18 $\underbrace{99 \cdots 9}_{n \uparrow \uparrow \uparrow \uparrow} 62=2 \times 94 \underbrace{99 \cdots 9}_{n-1} 81=2 \times(95 \underbrace{00 \cdots 0}_{n+10}-19)=2 \times 19 \times(5 \underbrace{00 \cdots 0}_{n+1}-1)$
Therefore, $p 、 q \quad r$ are $2 、 19 、 4 \underbrace{99 \cdots 9}_{n+1 \text { 9s }}$,
So $f(p)+f(q)+f(r)-f(p q r)=2+1+4+9 n+()-(9+9 n+8)=8$