Note that the result will be character.
H can have more than 4 characters, and will be interpreted 4 characters at a time from the beginning . So if H is not an exact multiple of 4 characters, this particular algorithm will not implicitly pad from the begining with 0's -- only the last group will be 0 padded.
This was interpretered as 01A9 (0)876.
If you needed to be padded with leading 0s implicitly then that could certainly be programmed.