Consider the following LC$-3$ code fragment $($the hex values in the first column give the address to which the corresponding instruction is loaded$)$: x$34$FE loop$1$ ADD R$4,$ R$5,$ R$6.....$ x$35$A$0$ BRnp loop$1$ ; NOTE THE TEST IS $"$np$"$ Given: the BR opcode is $0000$ NZP What does the instruction at x$35$A$0$ assemble to$?($Remember to account for the current value of the PC when the BR instruction is being executed, and the direction of the "jump", and the condition codes to be tested$)$ a$.$ x$0$B$5$D b$.$ x$0$B$5$E c$.$ x$0$B$5$F d$.$ x$0$AA$1$ e$.$ x$0$AA$2$ f$.$ x$0$AA$3$