Problem Statement

There is 1 pile with $N$ stones stacked in it.
A and B play a game in which they take turns splitting stones.
When splitting, a player divides stones into $2$ piles or $3$ piles.

Let $x$ be a positive integer. Then,

$2x$ stones can be split into $x$ stones and $x$ stones.
$2x+1$ stones can be split into $x$ stones and $x+1$ stones.
$3x$ stones can be split into $x$ stones, $x$ stones, and $x$ stones.
$3x+1$ stones can be split into $x$ stones, $x$ stones, and $x+1$ stones.
$3x+2$ stones can be split into $x$ stones, $x+1$ stones, and $x+1$ stones.

A player may choose any of the above operations.

For example, when there is initially $1$ pile with $5$ stones stacked in it:

Case 1:
Suppose A first splits it into $2$ stones, $2$ stones, and $1$ stone.
B splits a pile of $2$ stones, resulting in $2$ stones, $1$ stone, $1$ stone, and $1$ stone.
A also splits a pile of $2$ stones, resulting in $1$ stone, $1$ stone, $1$ stone, $1$ stone, and $1$ stone.
B cannot split a pile of $1$ stone any further.

Case 2:
Suppose A first splits it into $3$ stones and $2$ stones.
B can split the pile of $3$ stones into $2$ stones and $1$ stone, resulting in $2$ stones, $2$ stones, and $1$ stone.
A splits a pile of $2$ stones, resulting in $2$ stones, $1$ stone, $1$ stone, and $1$ stone.
B also splits a pile of $2$ stones, resulting in $1$ stone, $1$ stone, $1$ stone, $1$ stone, and $1$ stone.
A cannot split a pile of $1$ stone any further.

The player who becomes unable to split stones loses the game.
Therefore, in this game starting with $5$ stones,
A, who moves first, can always win by first splitting the stones into 3 piles, as in Case 1.
When A moves first and both A and B play optimally to win,
determine from the initial N whether A wins or B wins.

Input

N

The number of stones \( N(1 \le N \le 100) \) is given.

Output

Print the name of the winner, \( A \) or \( B \).
Print a newline at the end.

Sample

2

A

1

B

5

A

20

A