Bag of peanuts P65688

You have several peanuts inside a bag. Before you start eating some of them, you decide that you will eat exactly t peanuts in total. Repeatedly, you will take a peanut at random from the bag, and eat it. However, it happens that some of the peanuts are not complete, but just a half-peanut. Therefore, it is possible that you will not eat exactly t peanuts.

For instance, suppose that the bag has c = 1 complete peanuts, h = 2 half-peanuts, and that you want to eat exactly one peanut (that is, t = 1). In this case, with probability 1/3 you will eat the complete peanut, and stop. Otherwise, after eating a half-peanut, you will eat another peanut, which can be the remaining half-peanut (this would be a success, since you would have eaten 1/2 + 1/2 = t peanuts) or the complete peanut (this would be a failure, bacause you would have eaten 1/2 + 1 > t peanuts). Altogether, the probability of success is 1/3 + (2/3) · (1/2) = 2/3.

Given c, h and t, can you compute the probability of success?

Input

Input consists of several cases, with only integer numbers, each one with c, h and t. Assume 0 ≤ c ≤ 1000, 0 ≤ h ≤ 2000, and 0 ≤ t ≤ c + ⌊ h/2 ⌋.

Output

For every case, print with four digits after the decimal point the probability of eating exactly t peanuts when you are given a bag with c complete peanuts and h half-peanuts.

Hint

The expected solution has cost O(t). The given bounds for c, h and t are rather small, in order to reduce the magnitude of numerical errors. Even so, use the type long double and try hard to avoid underflows and overflows. Good luck!