<|im_start|>system
You are a helpful assistant. You first think about the reasoning process in the mind and then provide the user with the answer. You should try to truely solve the question by yourself without cheating.  <|im_end|>
<|im_start|>user
 Given a sequence $\{a_n\}$ whose sum of the first $n$ terms is $S_n$, and for any positive integer $n$, it holds that $2S_n=6-a_n$. The sequence $\{b_n\}$ satisfies $b_1=2$, and for any positive integer $n$, $b_{n+1}-b_n=2\log_{\frac{1}{3}}\left(\frac{a_n}{18}\right)$. If the sum of the first $n$ terms of the sequence $\left(\frac{1}{b_n}\right)$, denoted as $T_n < m$, always holds for all $n\in\mathbb{N}^*$, then the minimum value of the real number $m$ is \_\_\_\_\_. Show your work in <think> </think> tags. And return the final answer in <answer> </answer> tags.<|im_end|>
<|im_start|>assistant
Let me solve this step by step.
<think>
user
1
Given a sequence $\{a_n\}$ whose sum of the first $n$ terms is $S_n$, and for any positive integer $n$, it holds that $2S_n=6-a_n$. The sequence $\{b_n\}$ satisfies $b_1=2$, and for any positive integer $n$, $b_{n+1}-b_n=2\log_{\frac{1}{3}}\left(\frac{a_n}{18}\right)$. If the sum of the first $n$ terms of the sequence $\left(\frac{1}{b_n}\right)$, denoted as $T_n < m$, always holds for all $n\in\mathbb{N}^*$, then the minimum value of the real number $m$ is \_\_\_\_\_.
test