A little guess on it, the 10 letters b, d, k, l, r, v, w, x, y, z do not appear in this character set, therefore, among the 2048 words in BIP-39, cut the first 4 letters out first, and then delete all the words containing these 10 letters, leaving only about 600 candidate words.
 On the other hand, to uniquely express 12 seed words, each taking the first 4 letters requires 48 letters. Only 46 are given here. Therefore, there are, and only 2 words that are 3 letters long, and there are 10 words are 4 letters in length.
 If this guess is correct, it should be feasible to run a letter counting program for random combinations of 12 words on the compiled 600-length word list, but this may cause collisions, different combinations may produce the same letter count.
FYI.