Finding a grammar for $L=\{a^nb^mc^rd^s| n+m<r+s\}$

I am trying to find a grammar for $$L=\{a^nb^mc^rd^s| n+m, which has the hint of it having "some similarity" to $$L=\{a^ib^j|i

This last one is quite easy to get ($$S\to aSb | Sb | b$$), but still I am unsure on how to proceed.

I expect to manipulate the language by differentiating cases, and for example I found that if $$n>m$$, I will have to solve $$a^{m+k}b^mc^rd^s, k\ge1$$, and if $$n, I will have to solve $$a^nb^{n+k}c^rd^s, k\ge1$$, but, still I don't see the recursivity.

Any word $$w$$ in $$L$$ can be split into two words $$w$$ and $$w'$$ such that $$w \in \{a^n b^m c^r d^s \mid n + m = r+ s \}$$, $$w' \in \{c^r d^s \mid r+s > 0\}$$ and if the last symbol of $$w$$ exists and is $$d$$, then the first symbol of $$w$$ is $$d$$. On the other hand, every choice of $$w$$ and $$w'$$ that satisfies the above constrains induces a word $$w w' \in L$$.
In the following grammar the first two productions take care of generating $$w'$$, while the other productions generate $$w$$. In particular the nonterminal $$X_\alpha^\beta$$ with $$\alpha \in \{a,b\}$$ and $$\beta \in \{cd\}$$ will generate all the words in $$w \in \{a^n b^m c^r d^s \mid n + m = r+ s \}$$ such that (i) if $$\alpha=b$$: then $$n=0$$ and (ii) if $$\beta=c$$ then $$s=0$$.
\begin{align*} S &\to Sd \mid X_a^dd \mid Yc \mid X_a^cc \\ Y &\to Yc \mid X_a^cc \\ X_a^d &\to aX_a^d d \mid X_a^c \mid X_b^d \\ X_a^c & \to aX_a^cc \mid X_b^c \\ X_b^d &\to b X_b^d d \mid X_b^c \\ X_b^c &\to bX_b^cc \mid \varepsilon \end{align*}