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Description

Input

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.

Output

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.

Samples

3
9
11
2

3
1 0 2 
3
1 1 2
1
0

Note

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 $1$.