Add Candies
Description
There are $n$ bags with candies, initially the $i$-th bag contains $i$ candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
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Choose $m$ such that $1 \le m \le 1000$
Perform $m$ operations. In the $j$-th operation, you will pick one bag and add $j$ candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
It can be proved that for the given constraints such a sequence always exists.
You don't have to minimize $m$.
If there are several valid sequences, you can output any.
Each test contains multiple test cases.
The first line contains the number of test cases $t$ ($1 \le t \le 100$). Description of the test cases follows.
The first and only line of each test case contains one integer $n$ ($2 \le n\le 100$).
For each testcase, print two lines with your answer.
In the first line print $m$ ($1\le m \le 1000$) — the number of operations you want to take.
In the second line print $m$ positive integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le n$), where $a_j$ is the number of bag you chose on the $j$-th operation.
Input
Each test contains multiple test cases.
The first line contains the number of test cases $t$ ($1 \le t \le 100$). Description of the test cases follows.
The first and only line of each test case contains one integer $n$ ($2 \le n\le 100$).
Output
For each testcase, print two lines with your answer.
In the first line print $m$ ($1\le m \le 1000$) — the number of operations you want to take.
In the second line print $m$ positive integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le n$), where $a_j$ is the number of bag you chose on the $j$-th operation.
Samples
2
2
3
1
2
5
3 3 3 1 2
Note
In the first case, adding $1$ candy to all bags except of the second one leads to the arrangement with $[2, 2]$ candies.
In the second case, firstly you use first three operations to add $1+2+3=6$ candies in total to each bag except of the third one, which gives you $[7, 8, 3]$. Later, you add $4$ candies to second and third bag, so you have $[7, 12, 7]$, and $5$ candies to first and third bag — and the result is $[12, 12, 12]$.