Why is the following language  decidable? $L_{one\ right\ and\ never\ stops}$ I can't understand how the following language can ever be decidable:
$L= \{ \langle M \rangle |  M \ is \ a \ TM \ and\ there\ exists\ an\ input\ that\ in\ the\ computation\ $ $of\ M(w)\ the\ head\ only\ moves\ right\ and\ M\ never\ stops \}$, 
but apparently it is. 
We need to approve that there's at least one input as requested, so even if run on all inputs in parallel, we can't say something as "if it never stops, accept", it's paradox, and even if we could do that it was showing that the language is in $RE$- recursive enumerable.
so how come it is recursive?
Thanks a lot

As sdcvvc hints, this can be done due to how restricted the TMs in L are. The problem can be decided by examining the set of rules of the TM, which is finite. Firstly check if for some state $q_n$ the TM moves right on a blank input. This is the only way to go right infinitely as the tape contains only a finite number of symbols, other than a blank one. Secondly check that $q_n$ can be reached from $q_0$. For a normal TM this would not be decidable, but since this one can only go right, all of its input is the initial symbols of the tape.  
Consider a graph representing this TM. The vertices are states and edges are symbols on which state changes. The direction is always right and the written symbol is irrelevant, so they do not need to be represented. This problem is equivalent to first finding a cycle that only uses blank symbol edges and then a path from starting vertex to this cycle in the graph.
