Lorentz invariant measure, Scaling problem?

I’m having some trouble understanding some of the Fourier stuff in QFT.
Let’s say we adopt the following convention:
$$tilde{phi}(p) = int d^4x ,, e^{-ipx} phi(x) $$
which gives us the following inverse formula:
$$phi(x) = int frac{1}{(2pi)^4} d^4p,, e^{ipx} tilde{phi}(p) $$

Now in reality, the components of $p$ are not independent and are related buy the condition $omega^2 – vec{p}-m^2=0$. So we want to write:
$$phi_t(vec{x}) = int frac{1}{(2pi)^4} d^3vec{p},, domega,, e^{ivec{p}vec{x}} tilde{phi} (p) ,, delta({omega^2 – vec{p}-m^2}) $$

The goal is enforce the condition $omega^2 – vec{p}-m^2=0$ while still being Lorentz invariant and this is indeed achieved by adding this Dirac function.

But doing this won’t give me the right final expression, because I still end up with $frac{d^3vec{p}}{(2pi)^4} $ instead of $frac{d^3vec{p}}{(2pi)^3}$. I can’t get rid of the additional $2pi$ factor.

The problem of $delta({omega^2 – vec{p}-m^2})$ is that it is in a way arbitrary, one could use $delta({frac{1}{c}(omega^2 – vec{p}-m^2}))$ instead which also enforces the same condition. And in this case we end up with $frac{cd^3vec{p}}{(2pi)^4} $.

Since there is one correct answer, how should we choose the scaling ? and why, what’s the motivation? I feel that the scaling should be coherent with some other convention that we use but I’m not sure which one.