User:Spundun/Math/AP2

AP2

Arithmatic and geometric Progressions are 2 of the well known progressions in maths.

Arithmatic progression(AP) is a set in which the difference between 2 numbers in constant. for eg, 1,3,5,7,9 .... In this series the difference between 2 numbers is 2.

 

The task here is very simple indeed.

You will be given the 3rd term , 3rd last term and the sum of the series. You need print length of the series & the series.

Input

First line will contain a number indicating the number of test cases.

Each of the following t lines will have 3 number '3term' ,'3Lastterm' and 'sum'

3term  - is the 3rd term in of the series and

3Lastterm  - is the 3rd term in of the series and

sum - is the sum of the series.

Output

For each input of the test case, you need to print 2 lines.

fist line should have 1 value- number of terms in the series.

2nd line of the output should print the series numbers separated by single space.

Example

Input:
1
3 8 55

Output:
10
1 2 3 4 5 6 7 8 9 10

NOTE - In all the test cases all the series elements are positive integers.
The series will have at least 7 elements.
number of test cases <=100.
All the numbers will fit in 64 bits(long long in C)

Solution

The series will be of $ax+b$ nature. Where $1\leq x\leq N$

The three numbers we have are

$a(3)+b=\alpha (1)$

$a(N-2)+b=\beta (2)$

$\sum _{x=1}^{N}ax+b=a{\frac {N^{2}+N}{2}}+bN=\gamma (3)$

Subtracting 1 from 2.

$aN-5a=\beta -\alpha (4)$

Subtracting 1 from 3.

$a{\frac {N^{2}+N}{2}}+bN-3aN-bN=\gamma -\alpha N(5)$

Taking 4 further.

$a={\frac {\beta -\alpha }{N-5}}(6)$

Simplifying 5

$a{\frac {N(N-5)}{2}}=\gamma -\alpha N$

Substituting a

${\frac {\beta -\alpha }{N-5}}{\frac {N(N-5)}{2}}=\gamma -\alpha N$

${\frac {N(\beta -\alpha )}{2}}=\gamma -N\alpha$

$N{\frac {\beta +alpha}{2}}=\gamma$

$N={\frac {2\gamma }{\beta +\alpha }}$

$a={\frac {\beta -\alpha }{N-5}}$

$b=\alpha -3a$