What will be the output of the following pseudocode?Integer p, q, r, s Set p=1, q = 1for (each r from 0 to 2 )    for (each s from -4 to -2 )        p = p + 2if(p > r)            Continue         End if         p = 1if(p > s)             Jump out of the loop         End if     End for End for Print p + qQuestion 83AnswerA.13B.24C.35D.20

The pseudocode starts by setting p and q to 1. Then it enters a nested loop where r ranges from 0 to 2 and s ranges from -4 to -2. In each iteration of the inner loop, p is incremented by 2. If p is greater than r, the loop continues without executing the rest of the code in the loop. If p is greater than s, the loop is exited immediately.

Let's go through the loops:

• For r = 0, s = -4, p = 1 + 2 = 3. Since p > r (3 > 0), it continues to the next iteration without resetting p to 1.
• For r = 0, s = -3, p = 3 + 2 = 5. Since p > r (5 > 0), it continues to the next iteration without resetting p to 1.
• For r = 0, s = -2, p = 5 + 2 = 7. Since p > r (7 > 0), it continues to the next iteration without resetting p to 1.

The loop for r = 0 ends here. Now, r = 1.

• For r = 1, s = -4, p = 7 + 2 = 9. Since p > r (9 > 1), it continues to the next iteration without resetting p to 1.
• For r = 1, s = -3, p = 9 + 2 = 11. Since p > r (11 > 1), it continues to the next iteration without resetting p to 1.
• For r = 1, s = -2, p = 11 + 2 = 13. Since p > r (13 > 1), it continues to the next iteration without resetting p to 1.

The loop for r = 1 ends here. Now, r = 2.

• For r = 2, s = -4, p = 13 + 2 = 15. Since p > r (15 > 2), it continues to the next iteration without resetting p to 1.
• For r = 2, s = -3, p = 15 + 2 = 17. Since p > r (17 > 2), it continues to the next iteration without resetting p to 1.
• For r = 2, s = -2, p = 17 + 2 = 19. Since p > r (19 > 2), it continues to the next iteration without resetting p to 1.

The loop for r = 2 ends here.

Finally, it prints p + q = 19 + 1 = 20.

So, the output of the pseudocode will be 20 (Option D).