3. (20 points) ISA. Given a MIPS program on the left. Assume the current program counter Spe=0x40001024 which is the address of the jal instruction and the stack pointer Ssp OXFFFFOONO. Also given is the corresponding memory snapshot on the right. Let's execute another 10 instructions (the jal included). please (1) fill the updated memory content on the right (2) write down the values of the following registers Sao Svo. Sra. Ssp. and Spe (Each instruction is 4-byte wide. You may not need to fill all the blank entries in the memory. 4 bytes PC O40001024 Now HA Ox40000004 Ox00000006 00000005 Ssp- Ox00000001 0x40001010 0x40001014 Ox40001018 Ox40001010 Ox40001020 Ox40001024 Ox40001028 0x4000102C li $v0, 1 Fact: beq $a0, $v0, Lret sub $sp,$sp, 8 sw Sra, 4($sp) 8W 40, 0 (53p) sub $a0, $a0, $v0 tal Fact lw $a0, 0($sp) lw $ra, 4 ($sp) mult a0, $v0 mflo Svo add $sp,$sp, 8 Lret: ti sra 0x40001034 Ox40001038 Low Adde 0x4000103C 4 bytes - High Adder 0x40000004 0x00000006 After 10 instructions Low Adder 3. (20 points) ISA. Given a MIPS program on the left. Assume the current program counter Spe=0x40001024 which is the address of the jal instruction and the stack pointer Ssp OXFFFFOONO. Also given is the corresponding memory snapshot on the right. Let's execute another 10 instructions (the jal included). please (1) fill the updated memory content on the right (2) write down the values of the following registers Sao Svo. Sra. Ssp. and Spe (Each instruction is 4-byte wide. You may not need to fill all the blank entries in the memory. 4 bytes PC O40001024 Now HA Ox40000004 Ox00000006 00000005 Ssp- Ox00000001 0x40001010 0x40001014 Ox40001018 Ox40001010 Ox40001020 Ox40001024 Ox40001028 0x4000102C li $v0, 1 Fact: beq $a0, $v0, Lret sub $sp,$sp, 8 sw Sra, 4($sp) 8W 40, 0 (53p) sub $a0, $a0, $v0 tal Fact lw $a0, 0($sp) lw $ra, 4 ($sp) mult a0, $v0 mflo Svo add $sp,$sp, 8 Lret: ti sra 0x40001034 Ox40001038 Low Adde 0x4000103C 4 bytes - High Adder 0x40000004 0x00000006 After 10 instructions Low Adder