Funny Hand

 MoEngage has 

3$3$ decks. Each deck consists of N$N$ cards, numbered from 1$1$ to N$N$. He draws out 1$1$ card from each deck randomly with each card having an equal probability of being drawn.

MoEngage drew cards numbered A$A$ and B$B$ from the decks 1$1$ and 2$2$ 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$3$ 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$P$, you need to print the value P⋅N$P\cdot N$. It can be shown that this value is an integer.

Input Format

• First line will contain T$T$, 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,$N,A,$ and B$B$ 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⋅N$P\cdot N$ on a separate line.

Constraints

• 1≤T≤1000$1\le T\le 1000$
• 3≤N≤1000$3\le N\le 1000$
• 1≤A,B≤N$1\le A,B\le N$

Sample Input 1 

3
4 2 3
6 2 1
5 2 5

Sample Output 1 

2
1
0

Explanation

Test case 1$1$: Drawing the cards 1$1$ or 4$4$ will result in a funny hand. Hence, the probability will be P=12$P=\frac{1}{2}$. The value P⋅N=12⋅4=2$P\cdot N=\frac{1}{2}\cdot 4=2$.

Test case 2$2$: Drawing the card 3$3$ will result in a funny hand. Hence, the probability will be P=16$P=\frac{1}{6}$. The value P⋅N=16⋅6=1$P\cdot N=\frac{1}{6}\cdot 6=1$.

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