It turns out that using the div operator to compute the modulo operation is slow!

We can use a math trick to optimize the modulo operator $(\%).$ Compilers use this trick a lot.

If we have $"$x $\%$ y$",$ and y is a power of $2,$ such as $2^$n$,$ the result will be the lower n bits of x$.$

Therefore, we can use the lower register byte access to efficiently implement modulo!

Using only the following instruction$($s$)$:

mov

Please compute the following:

rax $=$ rdi $\%256$

rbx $=$ rsi $\%65536$

We will now set the following in preparation for your code:

rdi $=0$xe$572$

rsi $=0$xe$4$a$54$e$4$a

Please give me your assembly in bytes $($up to $0$x$1000$ bytes$)$: