Sufficient Statistic for a family of distributions consisting of Poisson family and Bernoulli family

Suppose $(X_1, . . . ,X_n)$ is an i.i.d. sample from the distribution $f_{\theta,k}(x)$ , where $\theta \in (0, 1)$ and $k = 1, 2$ . Assume that $$f_{\theta, k}(x)=\begin{cases} \text{Poisson($\theta)$}, &\text{if $k=1$}.\\ \\ \text{Bernoulli($\theta$)}, & \text{if $k=2$}. \end{cases}$$ . Check if $T=\sum_{i=1}^nX_i$ is a sufficient statistic for this family. If not, then find a sufficient statistic for this family. $$$$ My Attempt to the solutions is as follows : I found that $$\mathbb{P}(X_1=x_1, ...,X_n=x_n|T=t) =\begin{cases} \frac{n!}{x_1!x_2!....x_n!}(\frac{1}{n})^t & \text{if} &\sum_{i=1}^nx_i=t, &X_1, ...., X_n \sim \text{Poisson}(\theta) \\ \\\frac{1}{n \choose t} &\text{if} &\sum_{i=1}^nx_i=t, &X_1, ...., X_n \sim \text{Bernoulli}(\theta) \end{cases}$$ So $T$ is not sufficient for this family. $$$$ Now we can write the joint density as $$f_{\theta, k}(x_1, ...., x_n)=\frac{e^{-n\theta}(\theta)^{\sum_{i=1}^nx_i}}{\prod_{i=1}^n(x_i)!}\textbf{1}(k=1)+(\theta)^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}\textbf{1}(0 \leq x_{(1)} \leq x_{(n)} \leq 1)\textbf{1}(k=2)$$ The indicator $\textbf{1}(0 \leq x_{(1)} \leq x_{(n)} \leq 1)$ is because the support in our case is $\chi=\mathbb{N} \cup 0$ . So by the Factorization Theorem we get that $T(X_1, ...., X_n)=(\sum_{i=1}^nX_i, \prod_{i=1}^n(X_i)!, X_{(1)}, X_{(n)})$ is a sufficient statistic for this family as we can take $g_{\theta, k}(T(x_1, ...., x_n))$ equal to the density and $h(x_1, ...., x_n)=1$ . $$$$ Now to find…