To solve this challenge, we can implement a queue using two https://hometownstorageunit.com/. The primary mechanism behind this is that one stack will be used for enqueuing elements (the enqueue stack) and the other for dequeuing elements (the dequeue stack). When we want to enqueue an element, we simply push it onto the enqueue stack. However, when we want to dequeue an element or retrieve the front element, we need to ensure that the elements are in the correct order (FIFO). If the dequeue stack is empty, we pop all elements from the enqueue stack and push them onto the dequeue stack, which reverses their order. Then, the top of the dequeue stack represents the front of the queue.

Here's the Python code to implement the queue using two stacks and process the queries as per the problem statement:

class QueueWithTwoStacks:
    def __init__(self):
        self.enqueue_stack = []
        self.dequeue_stack = []

    def transfer_elements(self):
        while self.enqueue_stack:
            self.dequeue_stack.append(self.enqueue_stack.pop())

    def enqueue(self, x):
        self.enqueue_stack.append(x)

    def dequeue(self):
        if not self.dequeue_stack:
            self.transfer_elements()
        return self.dequeue_stack.pop()

    def peek(self):
        if not self.dequeue_stack:
            self.transfer_elements()
        return self.dequeue_stack[-1]

# Read number of queries
q = int(input())

# Create a queue
queue = QueueWithTwoStacks()

# Process each query
for _ in range(q):
    query = input().split()
    query_type = int(query[0])

    if query_type == 1:
        # Enqueue operation
        x = int(query[1])
        queue.enqueue(x)
    elif query_type == 2:
        # Dequeue operation
        queue.dequeue()
    elif query_type == 3:
        # Print operation
        print(queue.peek())

Let's break down the sample input and output:

• We perform 10 queries on the queue.
• We enqueue 42, the queue state is [42].
• We dequeue the first element, queue state becomes [].
• We enqueue 14, the queue state is [14].
• We print the first element, output is 14.
• We enqueue 28, the queue state is [14, 28].
• We print the first element, output is 14.
• We enqueue 60, then 78, the queue state is [14, 28, 60, 78].
• We dequeue twice, removing 14 and then 28, the queue state becomes [60, 78].

The output as per the sample input would be:

14
14

This code handles all the operations efficiently. The enqueue operation is (O(1)), while the dequeue and peek operations are (O(n)) in the worst case but (O(1)) amortized because elements are only moved from the enqueue stack to the dequeue stack when the dequeue stack is empty.