Wrong application of optical theorem

I am trying to use optical theorem* ( given in box 24.2 in Quantum Field Theory and the Standard Model, M. Schwartz ). I am trying to calculate the imaginary part of this diagram for the scalar field:

But I am getting extra factor of 2 in the calculation. I have calculated the imaginary part using complex analysis, and also using the Cutkosky cutting rules. I have also checked some papers. I am quite sure that I am getting an extra 2 factor when I calculate using the optical theorem. I suspect that I might have misunderstood the formula (24.2 ).

Can anyone please verify my understanding of the formula?

Here it goes

( m is the mass of both the particles, s is a Mandelstam variable, $$\lambda$$ is the coupling constant ) :

$$E_{CM}$$ stands for the sum of energies of the two incident particles in the Center of Momentum frame. It should be equal to $$\sqrt{s}$$.

$$|\vec{p_i}|$$ stands for the magnitude of the four-momentum of either one of the incident particles. It comes out to be $$\sqrt{\frac{s}{4} -m^2}$$.

$$\Sigma_x \sigma(A->X)$$ is the sum of scattering cross section of all the possible diagrams at order $$\lambda$$. In this case there is just one term to be summed over. And it is $$(4\pi) \frac{\lambda^2}{64\pi^2s}$$ . Here $$4\pi$$ comes from the integration of total solid angle. Remaning is just the differential cross section $$\frac{d\sigma}{d\Omega}$$ for scattering at order $$\lambda$$ .

The total answer that I get is $$\frac{\lambda^2}{8\pi} \sqrt{\frac{1}{4} - \frac{m^2}{s}}$$.

I have spent quite some time on this but still don't understand what am I doing wrong here. By every other method, I am getting an answer half this value. I can show more calculations and more arguments if anyone asks for them. Thanks.

Update:

*: Here is the formula: $$Im M(A->A) = 2E_{CM}|\vec{p_i}|\Sigma_x \sigma(A->X)$$

It turns out that I was calculating the cross section incorrectly. $$\Sigma_x \sigma(A->X)$$ should come out to be $$(2\pi) \frac{\lambda^2}{64\pi^2s}$$ because I have to divide the phase space by 2 ( output particles are identical ).