Prove that there exists a $2021$-digit positive integer $\overline{a_1a_2\ldots a_{2021}}$, with all its digits being non-zero, such that for every $1 \leq n \leq 2020$ the following equality holds $$\overline{a_1a_2\ldots a_n} \cdot \overline{a_{n+1}a_{n+2}\ldots a_{2021}}=\overline{a_1a_2\ldots a_{2021-n}} \cdot \overline{a_{2022-n}a_{2023-n}\ldots a_{2021}}$$and all four numbers in the equality are pairwise different.