Show that a Field Extension is Unramified Using Transitivity

Let $K=\mathbb{Q(\sqrt{5})}, L=\mathbb{Q(\sqrt{7})}, M=\mathbb{Q(\sqrt{35})}$, and $KL=\mathbb{Q(\sqrt{5},\sqrt{7})}$. Show that $KL/M$ is unramified (i.e. every prime ideal of $M$ is unramified in $KL$).

A hint provided is to only consider the ramification indices of $2$,$5$, and $7$ in extensions $L/\mathbb{Q}, K/\mathbb{Q}, M/\mathbb{Q}$ and to apply transitivity.

Based on what I know, I need to consider the discriminant of each extension:

disc$(K)=5$, disc$(L)=28=2^2\cdot7$, disc$(M)=140=2^2\cdot5\cdot7$ where I am using the fact that $Q(\sqrt{D})$ has discriminant either

$D$ if $D \equiv 1$ (mod 4), or $4D$ if $D \equiv 3$ (mod 4).

I tried to do three different lattices: one for each prime divisor: $(2,5,7)$. However, I am finding that each discriminant is divisible by $5$ in the diagram for $5$.

Can somebody please explain how to solve this problem? I have been struggling very much with this problem for 4 days now.

Thanks in advance, any help is greatly appreciated.