We have a chessboard and we call a $1\times1$ square a room. A robot is standing on one arbitrary vertex of the rooms. The robot starts to move and in every one movement, he moves one side of a room. This robot has $2$ memories $A,B$. At first, the values of $A,B$ are $0$. In each movement, if he goes up, $1$ unit is added to $A$, and if he goes down, $1$ unit is waned from $A$, and if he goes right, the value of $A$ is added to $B$, and if he goes left, the value of $A$ is waned from $B$. Suppose that the robot has traversed a traverse (!) which hasn’t intersected itself and finally, he has come back to its initial vertex. If $v(B)$ is the value of $B$ in the last of the traverse, prove that in this traverse, the interior surface of the shape that the robot has moved on its circumference is equal to $|v(B)|$.