Eccoci.

In the hexadecimal number system numbers are represented using $16$ different digits: $$0,1,2,3,4,5,6,7,8,9,\mathrm A,\mathrm B,\mathrm C,\mathrm D,\mathrm E,\mathrm F.$$

The hexadecimal number $\mathrm{AF}$ when written in the decimal number system equals $10 \times 16 + 15 = 175$.

In the $3$-digit hexadecimal numbers $10\mathrm A$, $1\mathrm A0$, $\mathrm A10$, and $\mathrm A01$ the digits $0$, $1$ and $\mathrm A$ are all present.
Like numbers written in base ten we write hexadecimal numbers without leading zeroes.

How many hexadecimal numbers containing at most sixteen hexadecimal digits exist with all of the digits $0$, $1$, and $\mathrm A$ present at least once?
Give your answer as a hexadecimal number.

(A, B, C, D, E and F in upper case, without any leading or trailing code that marks the number as hexadecimal and without leading zeroes, e.g. 1A3F and not: 1a3f and not 0x1a3f and not $1A3F and not #1A3F and not 0000001A3F)