Getting 88bytes cipher output from 48bytes input in AES The other day I saw this Hybrid Cryptographic Algorithm in the Google and felt curious about it. I am trying to understand this Hybrid Algorithm. And hopefully its not so hard to understand whats going on until AES portion comes up. Here in this research paper the author gives 48bytes input to AES and gets 88byte output from AES. 

I hope the author won't mind that I copied a part of the image and gave it in here.


My question is, How is this possible? AES always gives cipher size that is divisible by 16. But 88byte is wired! Can anyone help me by explaining or giving a hint what might be the way to achieve this?

-Thanks.
I don't think it's useful spending time on trying to understand that paper, but if you look at the screenshots and compare to their "character counts", you see that they are counting base64 characters and including the padding characters in the count.

That means "88 characters" could be one IV + three blocks of AES output (512 bits in base 64 = 85.33 + 2.67 padding).