CF1942G Bessie and Cards

[Second Dark Matter Battle - Pokemon Super Mystery Dungeon](https://soundcloud.com/decabat/second-dark-matter-battle?in=decabat/sets/best-of-pokemon-super-mystery) ⠀ Bessie has recently started playing a famous card game. In the game, there is only one deck of cards, consisting of $ a $ "draw $ 0 $ " cards, $ b $ "draw $ 1 $ " cards, $ c $ "draw $ 2 $ " cards, and $ 5 $ special cards. At the start of the game, all cards are in the randomly shuffled deck. Bessie starts the game by drawing the top $ 5 $ cards of the deck. She may then play "draw $ x $ " cards from the hand to draw the next $ x $ cards from the top of the deck. Note that every card can only be played once, special cards cannot be played, and if Bessie uses a "draw $ 2 $ " card when there is only $ 1 $ card remaining in the deck, then she simply draws that remaining card. Bessie wins if she draws all $ 5 $ special cards. Since Bessie is not very good at math problems, she wants you to find the probability that she wins, given that the deck is shuffled randomly over all $ (a + b + c + 5)! $ possible orderings. It can be shown that this answer can always be expressed as a fraction $ \frac{p}{q} $ where $ p $ and $ q $ are coprime integers. Output $ p \cdot q^{-1} $ modulo $ 998\,244\,353 $ .

The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. Each test case contains three integers $ a $ , $ b $ , and $ c $ ( $ 0 \le a, b, c \le 2 \cdot 10^5 $ ) – the number of draw $ 0 $ cards, draw $ 1 $ cards, and draw $ 2 $ cards, respectively. It is guaranteed that the sum of $ a $ over all test cases does not exceed $ 2 \cdot 10^5 $ , the sum of $ b $ over all test cases does not exceed $ 2 \cdot 10^5 $ , and the sum of $ c $ over all test cases does not exceed $ 2 \cdot 10^5 $ .

For each test case, output a single integer — the probability that Bessie wins, modulo $ 998\,244\,353 $ .

In the first case, we have $ 1 $ of each type of "draw" card and $ 5 $ special cards. There are $ 30\,720 $ starting decks where Bessie will win by drawing the top $ 5 $ cards and $ 40\,320 $ starting decks in total. Thus, the probability of Bessie winning is $ \frac{30\,720}{40\,320} = \frac{16}{21} $ . One example of a winning starting deck is, top to bottom, 1. "Special", 2. "Draw $ 1 $ ", 3. "Special", 4. "Special", 5. "Draw $ 0 $ ", 6. "Draw $ 2 $ ", 7. "Special", 8. "Special". One example of a losing starting deck is: 1. "Special", 2. "Draw $ 1 $ ", 3. "Special", 4. "Special", 5. "Draw $ 0 $ ", 6. "Special", 7. "Special", 8. "Draw $ 2 $ ".