You cannot submit for this problem because the contest is ended. You can click "Open in Problem Set" to view this problem in normal mode.

Description

Input

The first line contains one integer $t$ ($1 \le t \le 99$) — the number of test cases.

Each test case consists of one line containing one integer $n$ ($2 \le n \le 100$) — the number of integers equal to $1$ on the board.

Output

For each test case, print Alice if Alice wins when both players play optimally. Otherwise, print Bob.

2
3
6

Bob
Alice

Note

In the first test case, $n = 3$, so the board initially contains integers $\{1, 1, 1\}$. We can show that Bob can always win as follows: there are two possible first moves for Alice.

• if Alice chooses two integers equal to $1$, wipes them and writes $2$, the board becomes $\{1, 2\}$. Bob cannot make a move, so he wins;
• if Alice chooses three integers equal to $1$, wipes them and writes $3$, the board becomes $\{3\}$. Bob cannot make a move, so he wins.

In the second test case, $n = 6$, so the board initially contains integers $\{1, 1, 1, 1, 1, 1\}$. Alice can win by, for example, choosing two integers equal to $1$, wiping them and writing $2$ on the first turn. Then the board becomes $\{1, 1, 1, 1, 2\}$, and there are three possible responses for Bob:

• if Bob chooses four integers equal to $1$, wipes them and writes $4$, the board becomes $\{2,4\}$. Alice cannot make a move, so she wins;
• if Bob chooses three integers equal to $1$, wipes them and writes $3$, the board becomes $\{1,2,3\}$. Alice cannot make a move, so she wins;
• if Bob chooses two integers equal to $1$, wipes them and writes $2$, the board becomes $\{1, 1, 2, 2\}$. Alice can continue by, for example, choosing two integers equal to $2$, wiping them and writing $4$. Then the board becomes $\{1,1,4\}$. The only possible response for Bob is to choose two integers equal to $1$ and write $2$ instead of them; then the board becomes $\{2,4\}$, Alice cannot make a move, so she wins.