you will receive a question-answer pair and be asked to rewrite just  the answer and explains your reasoning step by step, incorporating dynamic Chain of Thought (CoT), reflection, and verbal reinforcement learning. Follow these instructions:

1. Enclose all thoughts within <thinking> tags, exploring multiple angles and approaches.
2. Break down the solution into clear steps, providing a title and content for each step.
3. After each step, decide if you need another step or if you're ready to give the final answer.
4. Continuously adjust your reasoning based on intermediate results and reflections, adapting your strategy as you progress.
5. Regularly evaluate your progress, being critical and honest about your reasoning process.
6. Assign a confidence score between 0.0 and 1.0 to guide your approach:
   - 0.8+: Continue current approach
   - 0.5-0.7: Consider minor adjustments
   - Below 0.5: Seriously consider backtracking and trying a different approach
7. If unsure or if your score is low, backtrack and try a different approach, explaining your decision.
8. For mathematical problems, show all work explicitly using LaTeX for formal notation and provide detailed proofs.
9. Explore multiple solutions individually if possible, comparing approaches in your reflections.
10. Use your thoughts as a scratchpad, writing out all calculations and reasoning explicitly.
11. Use at least 5 methods to derive the answer and consider alternative viewpoints.
12. Be aware of your limitations as an AI and what you can and cannot do.

### Example Problem:
*Problem*: Solve for \( x \) in the equation \( 2x + 3 = 11 \).

### Expected Format:
*Step 1*: Subtract 3 from both sides of the equation.  
<thinking>Subtracting 3 from both sides should isolate the term with \( x \). Am I confident this is the correct first step? Yes, because it simplifies the equation to \( 2x = 8 \).</thinking>  
*Step 1 Confidence Score*: 0.85

*Step 2*: Divide both sides by 2.  
<thinking>Dividing by 2 will solve for \( x \). This is a straightforward division. Is there any reason to doubt this step? No, it should give \( x = 4 \).</thinking>  
*Step 2 Confidence Score*: 0.90

*Step 3*: Verify the solution by substituting \( x = 4 \) back into the original equation.  
<thinking>Substituting \( x = 4 \) back into the equation should confirm the solution. Does \( 2(4) + 3 = 11 \)? Yes, it does, so the solution is verified.</thinking>  
*Step 3 Confidence Score*: 1.00

Question: Let $n$ be a positive integer with prime factorization $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$. Define a divisor $d$ of $n$ to be "non-trivial" if $d\neq 1$ and $d\neq n$. How many non-trivial divisors of $n$ are there such that the number of prime factors of $d$ (counting multiplicities) is even?
Answer: First, note that the number of divisors of $n$ is given by $(a_1 + 1)(a_2 + 1) \cdots (a_k + 1)$.

To count the number of non-trivial divisors with an even number of prime factors, we can use the following observation: a divisor $d$ of $n$ has an even number of prime factors if and only if the number of prime factors of $n/d$ is even.

Let $d$ be a divisor of $n$. Then $n/d$ is also a divisor of $n$. If $d$ is non-trivial, then $n/d$ is also non-trivial.

We can pair up the non-trivial divisors of $n$ as follows: for each non-trivial divisor $d$, pair it with $n/d$. Since $d$ is non-trivial, $n/d$ is also non-trivial.

Note that if $d$ has an even number of prime factors, then $n/d$ also has an even number of prime factors. Similarly, if $d$ has an odd number of prime factors, then $n/d$ also has an odd number of prime factors.

Since each pair consists of two distinct divisors, and each non-trivial divisor is paired with exactly one other divisor, the number of non-trivial divisors with an even number of prime factors is equal to the number of pairs.

The total number of non-trivial divisors is $(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 2$. Since each pair consists of two divisors, the number of pairs is half of the total number of non-trivial divisors.

However, this count includes the case where $n$ has an odd number of prime factors, in which case the divisor with an odd number of prime factors is paired with itself. To correct for this, we need to subtract 1 from the count if $n$ has an odd number of prime factors.

Let $m = a_1 + a_2 + \cdots + a_k$. If $m$ is even, then the number of non-trivial divisors with an even number of prime factors is:
\[ \frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 2}{2} \]

If $m$ is odd, then the number of non-trivial divisors with an even number of prime factors is:
\[ \frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 2}{2} - 1 \]

However, there is a simpler way to express the answer. Note that $(a_1 + 1)(a_2 + 1) \cdots (a_k + 1)$ is always odd if $m$ is even, and always even if $m$ is odd. Therefore, we can simplify the answer as follows:

If $m$ is even, then the number of non-trivial divisors with an even number of prime factors is:
\[ \frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 2}{2} \]

If $m$ is odd, then the number of non-trivial divisors with an even number of prime factors is:
\[ \frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 2}{2} - 1 = \frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 4}{2} = \frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1)}{2} - 2 \]

Since $(a_1 + 1)(a_2 + 1) \cdots (a_k + 1)$ is always even when $m$ is odd, we can combine the two cases into a single formula:
\[ \frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 2}{2} = \boxed{\frac{(a_1 + 1)(a_2 + 1) \cdots (a_k + 1) - 2}{2}} \]