Textbook RSA uses moduli of form N$=$pq with p and q of the same bitlength. For efficiency reasons one can also consider three$-$prime RSA for which N$=$pqr with p$,$q and r of the same bitlength since the best factoring algorithms can not yet exploit this form.

Let p$=7,$q$=11,$r$=17$ and let N$=$pqr$=1309.$ Let a three$-$prime RSA public key be $($N$,$e$)=(1309,13)$ with corresponding private key d$=517.$ Use a natural extension of the fast decryption method stated in the lectures and the Chinese Remainder Theorem, for p$,$q and r$,$ in order to decrypt the ciphertext c$=1234.$

Note: your calculations should be based on the knowledge of p$,$q and r and use the Chinese Remainder Theorem.