CF1864H Asterism Stream

Bogocubic is playing a game with amenotiomoi. First, Bogocubic fixed an integer $ n $ , and then he gave amenotiomoi an integer $ x $ which is initially equal to $ 1 $ . In one move amenotiomoi performs one of the following operations with the same probability: - increase $ x $ by $ 1 $ ; - multiply $ x $ by $ 2 $ . Bogocubic wants to find the expected number of moves amenotiomoi has to do to make $ x $ greater than or equal to $ n $ . Help him find this number modulo $ 998\,244\,353 $ . Formally, let $ M = 998\,244\,353 $ . It can be shown that the answer can be expressed as an irreducible fraction $ \frac{p}{q} $ , where $ p $ and $ q $ are integers and $ q \not \equiv 0 \pmod{M} $ . Output the integer equal to $ p \cdot q^{-1} \bmod M $ . In other words, output such an integer $ y $ that $ 0 \le y < M $ and $ y \cdot q \equiv p \pmod{M} $ .

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 100 $ ). The description of the test cases follows. The only line of each test case contains one integer $ n $ ( $ 1 \le n \le 10^{18} $ ).

For each test case, output a single integer — the expected number of moves modulo $ 998\,244\,353 $ .

In the first test case, $ n\le x $ without any operations, so the answer is $ 0 $ . In the second test case, for $ n = 4 $ , here is the list of all possible sequences of operations and their probabilities: - $ 1\stackrel{+1}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{+1}{\longrightarrow}4 $ , the probability is $ \frac{1}{8} $ ; - $ 1\stackrel{\times 2}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{+1}{\longrightarrow}4 $ , the probability is $ \frac{1}{8} $ ; - $ 1\stackrel{+1}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{\times 2}{\longrightarrow}6 $ , the probability is $ \frac{1}{8} $ ; - $ 1\stackrel{\times 2}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{\times 2}{\longrightarrow}6 $ , the probability is $ \frac{1}{8} $ ; - $ 1\stackrel{+1}{\longrightarrow}2\stackrel{\times 2}{\longrightarrow}4 $ , the probability is $ \frac{1}{4} $ ; - $ 1\stackrel{\times 2}{\longrightarrow}2\stackrel{\times 2}{\longrightarrow}4 $ , the probability is $ \frac{1}{4} $ . So the expected number of moves is $ 4 \cdot \left(3 \cdot \frac{1}{8}\right) + 2 \cdot \left(2 \cdot \frac{1}{4} \right) =\frac{5}{2} \equiv 499122179 \pmod{998244353} $ . In the third test case, for $ n = 8 $ , the expected number of moves is $ \frac{137}{32}\equiv 717488133\pmod{998244353} $ . In the fourth test case, for $ n = 15 $ , the expected number of moves is $ \frac{24977}{4096} \equiv 900515847 \pmod{998244353} $ .