Question 22 18 marks] Here is the declaration of a struct for the nodes in a singly-linked list: typedef struct intnode f int value; struct intnode *next; intnode_t; Variables head, tail and p are declared this way: intnode t *head; intnode_t *tail; intnode_t *p = NULL; head and tail point to the first and last nodes in a singly-inked 3, in that order: list that contains the integers 1, 2, and 2 head tall When we "rotate" the nodes one position to the right, the tail node becomes the head node, and the second node becomes the tail node. In the linked list shown here, the node containing 3 becomes the node at the head of the rotated linked list, and the node containing 2 becomes the node at the tail of the linked list, as shown here: head tail (a) The diagrams in the answer booklet depict the step-by-step execution of six C statements that rotate this linked list. Each step corresponds to the execution of one statement; however, the statements have been replaced by ruled lines. For cach step, write the missing statement. (In other words, on each ruled line, write the C statement that transforms the linked list shown above the line into the linked list shown below the line.) [6 marks] b) Would your solution for part (a) work for any singly-linked list with a head pointer and a tail ointer? Ifnot, why not? 12 marks]