CF1348D.Phoenix and Science

Phoenix has decided to become a scientist! He is currently investigating the growth of bacteria.

Initially, on day $1$, there is one bacterium with mass $1$.

Every day, some number of bacteria will split (possibly zero or all). When a bacterium of mass $m$ splits, it becomes two bacteria of mass $\frac{m}{2}$ each. For example, a bacterium of mass $3$ can split into two bacteria of mass $1.5$.

Also, every night, the mass of every bacteria will increase by one.

Phoenix is wondering if it is possible for the total mass of all the bacteria to be exactly $n$. If it is possible, he is interested in the way to obtain that mass using the minimum possible number of nights. Help him become the best scientist!

凤凰决定成为一名科学家！他目前正在研究细菌的生长。

初始时，在第 $1$ 天，有一个质量为 $1$ 的细菌。

每天，一些细菌会分裂（可能为零个，也可能全部分裂）。当一个质量为 $m$ 的细菌分裂时，它会变成两个质量均为 $\frac{m}{2}$ 的细菌。例如，一个质量为 $3$ 的细菌可以分裂成两个质量均为 $1.5$ 的细菌。

此外，每晚，每个细菌的质量都会增加 $1$。

凤凰想知道：是否有可能使所有细菌的总质量恰好等于 $n$？如果可能，他希望用最少的夜晚数来达到该总质量。请帮助他成为最优秀的科学家！

The input consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($2 \le n \le 10^9$) — the sum of bacteria masses that Phoenix is interested in.

输入包含多个测试用例。第一行包含一个整数 $t$（$1 \le t \le 1000$），表示测试用例的数量。

每个测试用例的第一行包含一个整数 $n$（$2 \le n \le 10^9$），表示 Phoenix 所关注的细菌总质量。

For each test case, if there is no way for the bacteria to exactly achieve total mass $n$, print -1. Otherwise, print two lines.

The first line should contain an integer $d$ — the minimum number of nights needed.

The next line should contain $d$ integers, with the $i$-th integer representing the number of bacteria that should split on the $i$-th day.

If there are multiple solutions, print any.

对于每个测试用例，如果细菌无法恰好达到总质量 $n$，则输出 -1。否则，输出两行。

第一行应包含一个整数 $d$ —— 所需的最少天数。

下一行应包含 $d$ 个整数，其中第 $i$ 个整数表示第 $i$ 天应分裂的细菌数量。

如果有多种解法，输出任意一种即可。

In the first test case, the following process results in bacteria with total mass $9$:

• Day $1$: The bacterium with mass $1$ splits. There are now two bacteria with mass $0.5$ each.
• Night $1$: All bacteria's mass increases by one. There are now two bacteria with mass $1.5$.
• Day $2$: None split.
• Night $2$: There are now two bacteria with mass $2.5$.
• Day $3$: Both bacteria split. There are now four bacteria with mass $1.25$.
• Night $3$: There are now four bacteria with mass $2.25$.

The total mass is $2.25+2.25+2.25+2.25=9$. It can be proved that $3$ is the minimum number of nights needed. There are also other ways to obtain total mass 9 in 3 nights.

$ $

In the second test case, the following process results in bacteria with total mass $11$:

• Day $1$: The bacterium with mass $1$ splits. There are now two bacteria with mass $0.5$.
• Night $1$: There are now two bacteria with mass $1.5$.
• Day $2$: One bacterium splits. There are now three bacteria with masses $0.75$, $0.75$, and $1.5$.
• Night $2$: There are now three bacteria with masses $1.75$, $1.75$, and $2.5$.
• Day $3$: The bacteria with mass $1.75$ and the bacteria with mass $2.5$ split. There are now five bacteria with masses $0.875$, $0.875$, $1.25$, $1.25$, and $1.75$.
• Night $3$: There are now five bacteria with masses $1.875$, $1.875$, $2.25$, $2.25$, and $2.75$.

The total mass is $1.875+1.875+2.25+2.25+2.75=11$. It can be proved that $3$ is the minimum number of nights needed. There are also other ways to obtain total mass 11 in 3 nights.

$ $

In the third test case, the bacterium does not split on day $1$, and then grows to mass $2$ during night $1$.

在第一个测试用例中，以下过程可得到总质量为 $9$ 的细菌：

• 第 $1$ 天：质量为 $1$ 的细菌分裂。此时共有两个质量均为 $0.5$ 的细菌。
• 第 $1$ 夜：所有细菌的质量均增加 $1$。此时共有两个质量均为 $1.5$ 的细菌。
• 第 $2$ 天：没有细菌分裂。
• 第 $2$ 夜：此时共有两个质量均为 $2.5$ 的细菌。
• 第 $3$ 天：两个细菌均分裂。此时共有四个质量均为 $1.25$ 的细菌。
• 第 $3$ 夜：此时共有四个质量均为 $2.25$ 的细菌。

总质量为 $2.25+2.25+2.25+2.25=9$。可以证明，$3$ 是所需夜晚数的最小值。此外，也存在其他方式在 $3$ 个夜晚内达到总质量 $9$。

$ $

在第二个测试用例中，以下过程可得到总质量为 $11$ 的细菌：

• 第 $1$ 天：质量为 $1$ 的细菌分裂。此时共有两个质量均为 $0.5$ 的细菌。
• 第 $1$ 夜：此时共有两个质量均为 $1.5$ 的细菌。
• 第 $2$ 天：其中一个细菌分裂。此时共有三个细菌，质量分别为 $0.75$、$0.75$ 和 $1.5$。
• 第 $2$ 夜：此时共有三个细菌，质量分别为 $1.75$、$1.75$ 和 $2.5$。
• 第 $3$ 天：质量为 $1.75$ 的细菌和质量为 $2.5$ 的细菌分裂。此时共有五个细菌，质量分别为 $0.875$、$0.875$、$1.25$、$1.25$ 和 $1.75$。
• 第 $3$ 夜：此时共有五个细菌，质量分别为 $1.875$、$1.875$、$2.25$、$2.25$ 和 $2.75$。

总质量为 $1.875+1.875+2.25+2.25+2.75=11$。可以证明，$3$ 是所需夜晚数的最小值。此外，也存在其他方式在 $3$ 个夜晚内达到总质量 $11$。

$ $

在第三个测试用例中，该细菌在第 $1$ 天不发生分裂，并在第 $1$ 夜增长至质量 $2$。