#!/usr/bin/env python3
"""Fold Ё/ё → Е/е in a word list and de-duplicate — the dictionary prep for "Эрудит".

The Эрудит ruleset has no Ё tile and treats Е/Ё as one letter, so its dictionary must be
folded before the DAWG is built. Folding merges pairs like ёж/еж, hence the de-dup. Output
is sorted (Russian order over the 32 folded letters) and LF-separated.

Run:  python3 dictprep/fold_yo.py dictprep/russian/scrabble.txt > /tmp/ru_erudit_words.txt
"""
import sys

ORDER = {c: i for i, c in enumerate("абвгдежзийклмнопрстуфхцчшщъыьэюя")}  # 32 letters, no ё


def key(w):
    return [ORDER.get(c, 99) for c in w]


def main():
    src = sys.argv[1] if len(sys.argv) > 1 else "/dev/stdin"
    words = {line.strip().replace("ё", "е").replace("Ё", "Е") for line in open(src, encoding="utf-8")}
    words.discard("")
    sys.stdout.write("\n".join(sorted(words, key=key)) + "\n")


if __name__ == "__main__":
    main()