How to derive scaling factors quickly for convolution theorem?

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In case I use the following convention for Fourier transform:

$$\mathcal{F(g(t))}=G(\omega)=\int_{-\infty}^{\infty}g(t)e^{-j\omega t}dt$$

and $$\mathcal{F^{-1}}(G(\omega))=g(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)e^{j\omega t}dt$$

What should be the scaling factors $a$, $b$ and $c$ for the following relations?

$$\mathcal{F}(f.g)= a \mathcal{F}(f)*\mathcal{F}(g)$$

$$f*g= b\mathcal{F}^{-1}(\mathcal{F}(f).\mathcal{F}(g))$$

$$f.g=c\mathcal{F}^{-1}(\mathcal{F}(f)*\mathcal{F}(g))$$

I guess $a,b,c$ should either be $2\pi$ or $\frac{1}{2\pi}$, but I'm not very sure how to directly get them by comparing with the standard case given on Wikipedia.

P.S: Here $j$ stands for iota.