HMMT 2022 C9

Consider permutations $\left(a_{0}, a_{1}, \ldots, a_{2022}\right)$ of $(0,1, \ldots, 2022)$ such that

• $a_{2022}=625$,
• for each $0 \leq i \leq 2022, a_{i} \geq \frac{625 i}{2022}$,
• for each $0 \leq i \leq 2022,\left\{a_{i}, \ldots, a_{2022}\right\}$ is a set of consecutive integers (in some order).

The number of such permutations can be written as $\frac{a!}{b!c!}$ for positive integers $a, b, c$, where $b>c$ and $a$ is minimal. Compute $100 a+10 b+c$.