Did you watch the movie "Animal World"? There is an interesting game in this movie.

The rule is like traditional Stone-Paper-Scissors. At the beginning of the game, each of the two players receives several cards, and there are three types of cards: scissors, stone, paper. And then in each round, two players need to play out a card simultaneously. The chosen cards will be discarded and can not be used in the remaining part of the game.

The result of each round follows the basic rule: Scissors beat Paper, Paper beats Stone, Stone beats Scissors. And the winner will get $1$ point, the loser will lose $1$ point, and the points will not change in the case of a draw.

Now, Rikka is playing this game with Yuta. At first, Yuta gets $a$ Scissors cards, $b$ Stone cards and $c$ Paper cards; Rikka gets $a'$ Scissors cards, $b'$ Stone cards, $c'$ Paper cards. The parameters satisfy $a+b+c=a'+b'+c'$. And then they will play the game exactly $a+b+c$ rounds (i.e., they will play out all the cards).

Yuta's strategy is "random". Each round, he will choose a card among all remaining cards with equal probability and play it out.

Now Rikka has got the composition of Yuta's cards (i.e., she has got the parameters $a,b,c$) and Yuta's strategy (random). She wants to calculate the maximum expected final points she can get, i.e., the expected final points she can get if she plays optimally.

Hint: Rikka can make decisions using the results of previous rounds and the types of cards Yuta has played.

The first line contains a single number $t(1\leq t \leq 10^4)$.

For each testcase, the first line contains three numbers $a,b,c$ and the second line contains three numbers $a',b',c'(0 \leq a,b,c,a',b',c' \leq 10^9, a+b+c =a' + b' + c'> 0)$.

For each testcase, if the result is an integer, print it in a line directly.

Otherwise, if the result equals to $\frac{a}{b}(|\gcd(a,b)| = 1, b > 0,$ $a$ and $b$ are integers$)$, output "$a$/$b$" (without the quote) in a single line.