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!

Public test cases

**Input**

1 2 1 3 0 3 0 6 3 2 1 2 1000 2000 1000

**Output**

0.6667 1.0000 1.0000 0.3333 0.7500

Information

- Author
- Salvador Roura
- Language
- English
- Official solutions
- C++
- User solutions
- C++
- Event
- Catorzè Concurs de Programació de la UPC - Semifinal
- Date
- 2016-06-29