Connect3

Connect3 is a simplified version of Connect4. Two players, black and white, take turns putting stones of their own color on a $4 \times 4$ board. Each square is written as $(a, b)$, where $a$ is the row and $b$ is the column. The lower left square is $(1, 1)$ and the upper right square is $(4, 4)$.

On a turn a player chooses a column, and the stone falls to the lowest empty square of that column. Square $(3, 1)$ can be taken only after $(2, 1)$ and $(1, 1)$ are occupied.

The board of Connect3. Each square is written as $(a, b)$ with row $a$ and column $b$.

The game ends the moment three stones of the same color stand next to each other in a row, in a column, or on a diagonal, and the player of that color wins. While no such triple exists, play goes on.

Black moves first and puts a stone on square $(1, x)$. After that white and black alternate. The game ends when white puts a stone on square $(a, b)$, so the white stone on $(a, b)$ is the one that completes three white stones in a line. Let $s$ be the board at that moment. Count how many different boards $s$ are possible. The order in which the stones were placed does not matter, so two boards are the same only when every square holds the same thing, a black stone, a white stone, or nothing. If no game can end this way, the answer is 0.

The first line has one integer $x$ ($1 \le x \le 4$), the column of the first stone, which black puts on square $(1, x)$.

The second line has two integers $a$ and $b$ ($1 \le a \le 4$, $1 \le b \le 4$), the square $(a, b)$ where the last stone of the game is put.

Print one integer, the number of different final boards $s$.