How to prove the following homology group isomorphism?

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I am new in this part. Can anyone help me to solve this question? Thanks!

Let $K=K_1\cup K_2$. $K_1 \cap K_2$ is $r$-dimension. Here $K, K_1 $ and $K_2$ are all simplicial complex.

Then the homology group $H_q(K)$ is isomorphic to direct sum of $H_q(K_1)$ with $H_q(K_2)$ for any $q>r+1$.

It is obvious to have $Z_q(K_1)\oplus Z_q(K_2) \subset Z_q(K)$. But I don't know how to proceed since I think there can be some simplex crossing over both $K_1$ and $K_2$. Then it does not lie in both $K_1$ and $K_2$. I have no idea how to deal with such kind of situation.

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We have a Mayer-Vietoris sequence for simplicial complexes. Searching the site, a post provides the following reference text (the result can be found at page $77$). Now, in our context this gives a long exact sequence of the form

$$ \dots \xrightarrow{\partial_{q}} H_q(K_1 \cap K_2) \xrightarrow{\iota_q} H_q(K_1) \oplus H_q(K_2) \xrightarrow{j_q} H_q(K) \xrightarrow{\partial_q} H_{q-1}(K_1 \cap K_2) \xrightarrow{\iota_{q-1}} \dots \\ \dots \xrightarrow{\partial_0} H_0(K_1 \cap K_2) \xrightarrow{\iota_0} H_0(K_1) \oplus H_0(K_2) \xrightarrow{j_0} H_0(K) \xrightarrow{} 0. $$

Note that, sice $\dim K_1 \cap K_2 = r$, then $H_q(K_1 \cap K_2) = 0$ when $q > r$. Now if $q > r+1$, the sequence above gives

$$ 0 \rightarrow H_q(K_1) \oplus H_q(K_2) \xrightarrow{j_q} H_q(K) \rightarrow 0 $$

which is precisely to say that $H_q(K) \simeq H_q(K_1) \oplus H_q(K_2)$ via $j_q$.