Alice and Bob have joined a competitive ~\pi e~ game! Initially, there are ~X~ slices of good ~\pi e~, and ~Y~ slices of bad ~\pi e~. With Alice going first, they will take turns picking one of the slices of ~\pi e~ uniformly at random and eating it. If a player eats a good slice of ~\pi e~, they gain one point, and if they eat a bad slice of ~\pi e~, they lose one point. After each turn, they will bake a good/bad slice of ~\pi e~ to replace the type of ~\pi e~ that was just eaten. Then, they will bake ~Z~ additional slices of ~\pi e~, each with equal probability of being either good or bad.

Alice and Bob will spend the rest of eternity competing in this ~\pi e~ game! Determine the limit of the expected value of Alice's points minus Bob's points after ~n~ turns, as ~n~ approaches infinity. To ensure the integrity of your solution, there will be ~T~ testcases.

Constraints

~1 \le T \le 314\,159~
~X + Y \ge 1~

Subtask 1 [27%]

~0 \le X,Y \le 314~
~1 \le Z \le 2~

Subtask 𝓮 [31%]

~0 \le X,Y \le 314~
~1 \le Z \le 271~

Subtask π [42%]

~0 \le X,Y \le 314\,159~
~1 \le Z \le 271\,828~

Input Specification

The first line contains one integer, ~T~, the number of testcases.

The next ~T~ lines each contain three space-separated integers, ~X~, ~Y~, and ~Z~, describing each testcase. ~X~ is the initial number of good slices of ~\pi e~, ~Y~ is the initial number of bad slices of ~\pi e~, and ~Z~ is the number of randomly good/bad ~\pi e~ slices which are added after each turn. It is guarenteed that there is at least one slice of ~\pi e~ initially.

Output Specificiation

Output ~T~ lines, the ~i^{th}~ line containing one real number, the answer to the ~i^{th}~ testcase. Your answers will be considered correct if their absolute error is less than ~10^{-6}~.

Sample Input

3
1 0 2
2 4 3
5 2 4

Sample Output

0.7853981633974483
-0.20456854629336979
0.2687567584009584

Explanation for Sample

For the first testcase, it can be shown that the limit of the expected value approaches ~\frac{\pi}{4} = 0.7853981633974483\dots~

For the second testcase, it can be shown that the limit of the expected value approaches ~\frac{2}{3}\ln(2) - \frac{2}{3} = -0.20456854629336979\dots~

For the third testcase, it can be shown that the limit of the expected value approaches ~\frac{3\sqrt{2}}{4}\ln(1+\sqrt{2}) - \frac{3\sqrt{2}}{8}\pi + 1 = 0.2687567584009584\dots~