How to find the congruence relation for the summation $\sum_{\substack{m<n_1<\cdots<n_r<p \\ n_i \equiv i \pmod 2}}\frac{1}{n_1\ldots n_r}$

Question

Let $p$ be a prime number and $j\ge 0$. By changing the variables $n_i \mapsto p-n_i$, I know that $$\sum_{m<n_1<\cdots<n_r<p}\frac{1}{n_1\ldots n_r} \equiv (-1)^r \sum_{0<n_r< \ldots <n_1<p-m} \frac{1}{n_1\ldots n_r} \pmod p.$$

I want to know how the above congruence works when we have the summation $$\sum_{\substack{m<n_1<\cdots<n_r<p \\ n_i \equiv i \pmod 2}}\frac{1}{n_1\ldots n_r}$$