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 $0 \le y \lt M$ and $y \cdot q \equiv p \pmod{M}$.

Bogocubic 正在与 amenotiomoi 玩一个游戏。首先，Bogocubic 固定一个整数 $n$，然后他给 amenotiomoi 一个初始值为 $1$ 的整数 $x$。

在每一步中，amenotiomoi 以相同概率执行以下两种操作之一：

• 将 $x$ 加 $1$；
• 将 $x$ 乘以 $2$。

Bogocubic 想要求出 amenotiomoi 使得 $x$ 大于或等于 $n$ 所需的期望步数。请帮助他求出该期望值对 $998\,244\,353$ 取模的结果。

形式化地，令 $M = 998\,244\,353$。可以证明，答案可表示为既约分数 $\frac{p}{q}$，其中 $p$ 和 $q$ 为整数，且 $q \not \equiv 0 \pmod{M}$。请输出满足 $p \cdot q^{-1} \bmod M$ 的整数。换言之，输出满足 $0 \le y < M$ 且 $y \cdot q \equiv p \pmod{M}$ 的整数 $y$。

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 $n$ ($1 \le n \le 10^{18}$).

每个测试包含多个测试用例。第一行包含测试用例的数量 $t$（$1 \le t \le 100$）。随后是各测试用例的描述。

每个测试用例仅有一行，包含一个整数 $n$（$1 \le n \le 10^{18}$）。

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

对于每个测试用例，输出一个整数——即期望移动次数对 $998\,244\,353$ 取模的结果。

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 $\frac{24977}{4096} \equiv 900515847 \pmod{998244353}$.

在第一个测试用例中，无需任何操作即有 $n\le x$，因此答案为 $0$。

在第二个测试用例中，当 $n = 4$ 时，以下是所有可能的操作序列及其对应概率：

• $1\stackrel{+1}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{+1}{\longrightarrow}4$，概率为 $\frac{1}{8}$；
• $1\stackrel{\times 2}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{+1}{\longrightarrow}4$，概率为 $\frac{1}{8}$；
• $1\stackrel{+1}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{\times 2}{\longrightarrow}6$，概率为 $\frac{1}{8}$；
• $1\stackrel{\times 2}{\longrightarrow}2\stackrel{+1}{\longrightarrow}3\stackrel{\times 2}{\longrightarrow}6$，概率为 $\frac{1}{8}$；
• $1\stackrel{+1}{\longrightarrow}2\stackrel{\times 2}{\longrightarrow}4$，概率为 $\frac{1}{4}$；
• $1\stackrel{\times 2}{\longrightarrow}2\stackrel{\times 2}{\longrightarrow}4$，概率为 $\frac{1}{4}$。

因此，操作步数的期望值为 $4 \cdot \left(3 \cdot \frac{1}{8}\right) + 2 \cdot \left(2 \cdot \frac{1}{4} \right) =\frac{5}{2} \equiv 499122179 \pmod{998244353}$。

在第三个测试用例中，当 $n = 8$ 时，操作步数的期望值为 $\frac{137}{32}\equiv 717488133\pmod{998244353}$。

在第四个测试用例中，当 $n = 15$ 时，操作步数的期望值为 $\frac{24977}{4096} \equiv 900515847 \pmod{998244353}$。