We construct a series of counterexamples disproving the long standing hypothesis stating that for two circle diffeomorphisms with breaks to be smoothly conjugate it is necessary to be `break-equivalent’ (in particular, that for a single circle diffeomorphism with breaks to be smoothly linearizable it is necessary for its breaks to compensate each other along every particular trajectory). The example constructed is a piecewise linear circle homeomorphism that has four break points lying on four distinct trajectories, and whose invariant measure is absolutely continuous w. r. t. the Lebesgue measure (but not piecewise C1). The irrational rotation number for such a map can be chosen to be a Roth number, but not that of bounded type.