Solving integral with absolute value

With a given $$x > 0$$ (I think we could restrict it to $$x \in [0, 3]$$), I'm trying to find the following integral:

$$\int_{z = 0}^{\min(x,1)} \int_{y = 0}^{\min(x - z, 1)} |x -z -y -1| dydz$$

However, here I'm not sure how to deal with absolute values. By the construction, the value under the modulus seems to be almost always negative when $$x < 1$$, and it feels like negative when $$x \leq 3$$, but I feel like I'm missing something.

Any hints on how should we proceed further?

Like you said, the hardest part here is the absolute value. We can make our lives easier by defining $$u := x-z-1$$, which simplifies the integrand into $$|u - y|$$. Since this function has a constant analytic form in the regions $$y \leq u$$ and $$y \geq u$$, we can rewrite our integral as the piecewise function:
$$\int_{0}^{\min(x, 1)}\left[ \int_{0}^{\min(u, 1)}(u-y)\ dy + \int_{\max(u, 0)}^{\min(1+u, 1)}(y-u)\ dy \right]\ dz$$
You still have to deal with the differing behavior of these integrals for different values of $$u$$, but this should be more mechanical than directly reasoning about the original absolute value. Just use the predictable behavior of $$\min(a,b)$$ and $$\max(a,b)$$, along with the fact that $$\int_{a}^{b}f(y)\ dy = 0$$ when $$a\geq b$$, as seen by the fact that the region of integration $$[a, b] := \left\{\ x\ |\ a\leq x\leq b \right\}$$ is empty.
After solving the inner integral for each separate region of $$u$$ (namely, $$u\leq0$$, $$u\geq1$$, and $$0\leq u\leq1$$), you just replace all your $$u$$'s by $$x-z-1$$ (including in your inequalities defining the different regions), and solve the outer integral. This requires more of the same case-by-case reasoning, but by this point you'll have the hang of it :)