In addition to the standard ways of dealing with this given by other commenters (keeping the original input around), perhaps more interesting is to imagine how befreak might compute 12 mod 4 using repeated subtraction.
I'm not clever enough to write the solution, but I imagine a loop using the branching operators (<, >, v, ^). Since each time through the loop pushes a bit onto the control stack, you have a built-in count for the number of times the loop was traversed.
In the example of 12 mod 4, the program would loop through 3 times and break once giving you 0001 (or 1110) on the control stack and something like 12, 8, 4, 0 on the stack. Then when reversing, control bits would be popped off and you'd end up going back 0, 4, 8, 12.
I'm not clever enough to write the solution, but I imagine a loop using the branching operators (<, >, v, ^). Since each time through the loop pushes a bit onto the control stack, you have a built-in count for the number of times the loop was traversed.
In the example of 12 mod 4, the program would loop through 3 times and break once giving you 0001 (or 1110) on the control stack and something like 12, 8, 4, 0 on the stack. Then when reversing, control bits would be popped off and you'd end up going back 0, 4, 8, 12.