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.

Input

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 $q$ ($1 \le p, q \le 10^{18}$).

Output

For each input case, output:

• "Alice" if Alice wins;
• "Bob" if Bob wins.

Note

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.