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.