The two versions differ in the limits, including variable bounds, as well as time and memory constraints. You can hack only if you solve both versions of this problem. Note that a correct solution for the hard version is not necessarily a valid solution for the easy version. We have a control car that behaves like a random variable. As a random variable is neither truly random nor truly variable, our control car is neither a car nor controllable! Anyhow, we are given a grid of size (n \times m) formed by the intersections of (n+1) horizontal lines and (m+1) vertical lines. The grid contains ((n+1)(m+1)) intersection points, which are the corners surrounding the cells. An orientation is an assignment of one of the four cardinal directions (up, down, left, right) to each of these ((n+1)(m+1)) intersection points. Therefore, there are (4^{(n+1)(m+1)}) possible orientations. Consider the process below for an orientation: For each point in the grid, we draw a wall extending in the direction it is oriented with length one unit. Walls may overlap or extend beyond the grid boundary. Place the control car at the top-left corner cell ((1, 1)). Then, as long as the car has not stopped, it moves according to these rules: If it can move down (i.e., no wall blocks the path to the cell directly below), it moves down because of gravity. If it cannot move down but can move right (i.e., no wall blocks the path to the cell directly to its right), it moves right. If the car exits the grid or cannot move, it stops. If it can move down (i.e., no wall blocks the path to the cell directly below), it moves down because of gravity. If it cannot move down but can move right (i.e., no wall blocks the path to the cell directly to its right), it moves right. If the car exits the grid or cannot move, it stops. The orientation is valid if the control car stops inside the grid. Count the number of valid orientations. As the number of valid orientations can be h