Let $T = \{9^k : k ~ {is ~an~ integer}, 0 \le k \le 4000\}$ . Given that $9^{4000}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T$ have 9 as their leftmost digit?

Let's think step by step and output the final answer within \boxed{}.
user