# $N$-body phase space for Fermi golden rule

I was following along Mark Thomson's Modern Particle Physics, and stumble upe the derivation of d$$n$$ of Fermi golden rule on page 62:

"... For the decay of a particle to a final state consisting of $$N$$ particles, there are $$N-1$$ independent momenta in the final state. Thus, the numer of independent states for an N-particle final state is $$$$dn=\prod^{N-1}_{i=1} dn_i = \prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3} \tag{A}.$$$$ This can be expressend in a more democratic form including the momentum space volume for the $$N$$th particle $$d^3\text{p}_N$$ and using delta-function to impose momentum conservation $$$$dn=\prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3} \delta^3\left(\textbf{p}_a-\sum^N_{i=1}\textbf{p}_i\right)d^3\textbf{p}_N \tag{B},$$$$ where $$\textbf{p}_a$$ is the momentum of the decaying particle.... (and so on) "

Based on the explanation between the steps, I don't understand why the three-dimensional dirac-delta function appear out of nowhere. I do know that physical significances imposing momentum conservation, but why does it have to take the form of dirac-delta? is there an intermediate steps that I'm missing here?

$$$$\int d^nx \delta^n(x-a) = 1\\$$$$ similiarly, $$$$\int d^3\textbf{p}_N \;\delta^3\left(\textbf{p}_a-\sum^N_{i=1} \textbf{p}_i\right) = 1$$$$ Therefore, from (A) \begin{align} dn&=\prod^{N-1}_{i=1} dn_i, \\ &= \prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3}, \\ &= \prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3} \int \delta^3\left(\textbf{p}_a-\sum^N_{i=1}\textbf{p}_i\right)d^3\textbf{p}_N. \end{align} can be obtained (B) with slight correction to the book.