Chain of Fools#2676
Many of you have heard the story of Turing's bicycle: It is said that the crank of his bicycle had a broken prong, and the chain had one bent link. When the bent link on the chain arrived at the same position as the broken prong, the chain would fall off and Turing would have to stop and put the chain back on. But being who he was, Turing could predict exactly when this would happen—that is, how many pedal strokes it would take—so he would dismount before it occurred, gently move the pedals by hand to let the undesired coupling pass, and then merrily (we imagine) continue on his way. (A picture of the sprocket-chain setup is shown below.)
Your task is to calculate the number of revolutions required for the first failure to occur in such a situation as Turing's. You will be given the number of prongs on the front sprocket, the number of links on the chain, the position of the broken prong, and the position of the bent link in the chain. The top prong is at position , the next prong forward is at position , up to prong number s-1. (See the diagram. Note that prong s-1 is the next prong to move to the top as Turing pedals.) The positions of the links are similar: the link at the top of the sprocket is at position , proceeding forward to c-1. The chain falls off when the broken prong and bent link are both at position .
Input for each test case consists of one line of the form s~c~p~l, where is the number of prongs on the front sprocket (1 < s < 100), is the number of links in the chain (200 > c > s), is the initial position of the broken prong, and is the initial position of the bent link. The last input line is .
The broken prong and bent link will never both start at position .
For each test case, output a single line in the following format:
Case n: r m/s
if it requires r~\frac{m}{s} revolutions for the first failure to occur, or
Case n: Never
if it can never happen.
The denominator of the fraction will always be the number of prongs on the sprocket; the fraction will not necessarily be reduced to lowest terms. Always print the values of and , even if .
Example 1