Expectation of absolute value of Brownian motion

Question

I'm working on this problem that I can't seem to figure out. The problem involves a 1-dimensional Brownian motion, $B_t$, where the subscript denotes the time, and it asks me to show that the expectation \begin{align} \mathbb{E}[|B_t-B_s|^4]=3|t-s|^2 \end{align} where $s<t$. I'm new to the subject of Brownian motion, so I'm not quite sure how to deal with this. I know that $B_t-B_s$ is a normally distributed random variable with mean zero and variance $t-s$. My intuition was therefore to compute \begin{align} \mathbb{E}[|B_t-B_s|^4] &= \int_{-\infty}^{\infty} |x|^4 ~f(x)~dx\\ &=\int_{-\infty}^{\infty} x^4~f(x)~dx \end{align} where $x$ denotes the value of the random variable $B_t-B_s$, and $f(x)$ denotes its distribution, $f(x)=\frac{1}{\sqrt{2\pi(t-s)}}e^{-\frac{x^2}{2(t-s)}}$. However, that's just the 4th moment of a normal distribution, which is $3(t-s)^4$, so I know that something is wrong with my intuition. Can someone point out what I'm doing wrong, or give a hint to what a better (or correct) approach would be?