Pseudorandom generator: Given a pseudorandom generator (PRG) G : {0, 1^}n {0, 1}²ⁿ , and let us denote G(x) = y₁||y₂, where |y₁| = |y₂| = n

Question: Let us define G : {0, 1}ⁿ {0, 1}⁴ⁿ as follows, on input an n-bit string x, First, it runs G(x) and obtains y₁||y₂;

Second, it runs G again on the first n-bit of the output y₁, i.e., computes G(y₁) and obtains z₁,...,z_2n. Then, it runs G again on the second n-bit of the output y₂, i.e., computes G(y2) and obtains z_2n+1,...,z_4n. Finally, outputs z₁, . . . , z_4n. Note that the final output will be 4n bits.

Is G() still a PRG? If yes, explain why; if not, give an explicit attack that violates the PRG definition, i.e., define a distinguisher explicitly.