The program execution of Figure 3.5 is described using six steps. Expand this description to show the use of the MAR and MBR. Memory 300 19 4 0 301 5 9 4 1 3022 9 4 1 CPU registers Memory 3 0 0 PC 300 1941 AC 301594 19 4 0 IR 302 2 9 4 CPU registers 3 0 1 PC 0 0 3 AC 1 9 4 0 IR 1 940 0 941 0 0 0 0 0 3 2 940 0 941 0 0 0 0 0 3 2 Step 1 Memory 300 19 4 0 301 5 9 4 302 2 9 4 CPU registers 30 PC 0 0 0 3 AC 59 4 1 IR Step 2 Memory 300 19 4 0 301 5 9 4 1 302 2 94 CPU registers 3 O 2 PC 0 0.0 5 AC 5 9 4 1 IR 1 940 0 941 0 Step 3 0 0 0 0 3 2 940 0 941 0 0 0 0 0 3 2 Step 4 Memory 300 | 1940 301 594 302 2 9 4 CPU registers 3 0 2 PC 0 0 0 S AC 2 9 4 1 IR Memory 300 1 9 4 301S 9 302 2 2 CPU registers 3 0 3 PC 0 0 0 5 AC 2 9 4 1 IR 3 940 0 941 0 Step 5 0 0 0 0 3 2 940 0 941 0 Step 6 0 0 0 0 Figure 3.5 Example of Program Execution (contents of memory and registers in hexadecimal)