codeforces#P1841A. Game with Board

Game with Board

Description

Alice and Bob play a game. They have a blackboard; initially, there are $n$ integers written on it, and each integer is equal to $1$.

Alice and Bob take turns; Alice goes first. On their turn, the player has to choose several (at least two) equal integers on the board, wipe them and write a new integer which is equal to their sum.

For example, if the board currently contains integers $\{1, 1, 2, 2, 2, 3\}$, then the following moves are possible:

  • choose two integers equal to $1$, wipe them and write an integer $2$, then the board becomes $\{2, 2, 2, 2, 3\}$;
  • choose two integers equal to $2$, wipe them and write an integer $4$, then the board becomes $\{1, 1, 2, 3, 4\}$;
  • choose three integers equal to $2$, wipe them and write an integer $6$, then the board becomes $\{1, 1, 3, 6\}$.

If a player cannot make a move (all integers on the board are different), that player wins the game.

Determine who wins if both players play optimally.

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.

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

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.