Help is appreciated, thank you!

Recall that the indicator function of rationals is not Riemann integrable. Prove directly (by constructing appropriate partitions) that the function f given below is Riemann integrable: f : [0, 1] - R, defined as f (0) = 0, f(x) = 0 for x irrational, and for rational x = k, f(x) = 1. (Here we assume that k and n are coprimes i.e. they dont have common prime factors.)