Solve the programming task below in a Python markdown code block.
MoEngage has 3 decks. Each deck consists of N cards, numbered from 1 to N. He draws out 1 card from each deck randomly with each card having an equal probability of being drawn.

MoEngage drew cards numbered A and B from the decks 1 and 2 respectively. Now, he wonders what is the probability that he will end up with a *funny hand* after drawing the third card.

A *funny hand* is when 3 consecutive numbered cards are present in your hand.  
Help MoEngage calculate the probability of ending up with a *funny hand* after drawing the third card.

If the final probability of ending up with a *funny hand* is P, you need to print the value P\cdot N. It can be shown that this value is an integer.

------ Input Format ------ 

- First line will contain T, the number of test cases. Then the test cases follow.
- Each test case contains a single line of input consisting of three space-separated integers N, A, and B denoting the size of each deck, the card drawn from the first deck, and the card drawn from the second deck respectively.

------ Output Format ------ 

For each test case, output a single integer, denoting P\cdot N on a separate line.

------ Constraints ------ 

$1 ≤ T ≤ 1000$
$3 ≤ N ≤ 1000$
$1 ≤ A, B ≤ N$

----- Sample Input 1 ------ 
3
4 2 3
6 2 1
5 2 5

----- Sample Output 1 ------ 
2
1
0

----- explanation 1 ------ 
Test case $1$: Drawing the cards $1$ or $4$ will result in a funny hand. Hence, the probability will be $P = \frac{1}{2}$. The value $P \cdot N = \frac{1}{2} \cdot 4 = 2$.

Test case $2$: Drawing the card $3$ will result in a funny hand. Hence, the probability will be $P = \frac{1}{6}$. The value $P \cdot N = \frac{1}{6} \cdot 6 = 1$.

Test case $3$: No matter what card MoEngage draws, he will never end up with a funny hand.