CF2196A Game with a Fraction

Alice and Bob have two integers $ p $ and $ q $ , and they are playing a game with these numbers. The players take turns, with Alice going first. On their turn, a player can do one of two actions: - decrease $ p $ by one (this action is possible if $ p \gt 0 $ ); - decrease $ q $ by one (this action is possible if $ q \gt 1 $ ). The game ends when $ p = 0 $ and $ q = 1 $ . Bob wins if at any point during the game the fraction $ \frac{p}{q} $ is equal to in value the fraction $ \frac{2}{3} $ . Otherwise, Alice wins. Given the initial values of $ p $ and $ q $ , determine the winner of the game if both players play optimally.

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 10^4 $ ). The description of the test cases follows. Each input case consists of a single line containing two integers $ p $ and $ q $ ( $ 1 \le p, q \le 10^{18} $ ).

For each input case, output: - "Alice" if Alice wins; - "Bob" if Bob wins.

In the first input case, the fraction is already equal to $ \frac{2}{3} $ by value, so Bob wins. In the second input case, one possible sequence of the game is as follows: - initially $ p = 10, q = 14 $ ; - after Alice's turn $ p = 9, q = 14 $ ; - after Bob's turn $ p = 9, q = 13 $ ; - after Alice's turn $ p = 9, q = 12 $ ; - after Bob's turn $ p = 8, q = 12 $ . Bob wins, as $ \frac{8}{12} $ is equal to $ \frac{2}{3} $ . It can be shown that in this example, with optimal play from both players, Bob always wins. For the third input case, Alice's optimal strategy will be to decrease $ q $ as long as possible. In this case, the game will end in favor of Alice regardless of Bob's actions.