If you have a 4bit string of binary, you got a possibly 16 numbers in decimal (0 through 15). Lets say each of these represents somehow, a character. We have 26 letters in our alphabet. If 1 of those 16 were a modifier for hte next 4bit chunk, you'de be left to only 15 letters on the first loop, and 16 on the next loop. For a total of 31 characters. Minus 26 for hte alphabet, leaves us with 5 characters to use as punctuation. ' , . ? ! (space) are the most common, which is 6 punctuation marks. I'm betting he didn't use one of the other punctuations in his encryption.
After breaking it all into 4bit chunks, I get this:
1100
0110
0111
1101
0000
0100
0000
0100
0101
0101
0010
0110
1000
0001
0011
1110
0010
1100
0001
0010
0000
0000
0010
0000
0001
1101
0100
0000
0000
0010
0000
1000
0011
0111
1100
0100
0111
0001
0000
0001
1100
0110
0101
1001
0000
1000
0000
1010
0000
1111
1111
0100
1000
0000
0010
0001
1000
0000
0110
1101
0000
1000
0000
0011
1110
0000
1110
0010
0100
0000
1100
0000
0010
0000
1100
1010
0111
1111
0000
0000
0000
1001
0100
It's unlikely that he doubled a modifier above a modifier, so 0000, 0101, and 1111 aren't it.
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