How does one decrypt the many time pad:
($R$ is a random string, $C$ is ciphertext, $P$ is plaintext)
$R_0 \oplus R_1 = C_0$
$R_2 \oplus R_1 = C_1$
$P_0 \oplus R_1 = C_2$
assuming that $R_n$ are perfectly random pads, plaintext is writing and $C_n$ are the encrypted pads. The key $R_1$ is used three times but it doesn't seem like Crib Dragging (the solution I've seen described to decrypt many time pads) would work (edit: to retrieve $R_0$, $R_2$, and $P_0$ or even just $P_0$) since the other values that $R_1$ is added to are random. Is there something I'm missing?