### Problem Statement

A number n taken to the falling factorial power k is defined as n*(n-1)*...*(n-k+1). We will denote it by n^^k. For example, 7^^3=7*6*5=210. By definition, n^^1=n.

We will now continue this definition to the non-positive values of k using the following fact: (n-k)*(n^^k)=n^^(k+1), or, in other words, n^^k=(n^^(k+1))/(n-k). It is directly derived from the above definition.

By using it, we find:

• n^^0=n^^1/(n-0)=1,
• n^^(-1)=n^^0/(n+1)=1/(n+1),
• n^^(-2)=1/(n+1)/(n+2),
• and, in general, n^^(-k)=1/(n+1)/(n+2)/.../(n+k).
For example, 3^^(-1)=1/4=0.25, 2^^(-3)=1/3/4/5=1/60=0.016666...

Given a positive int n and an int k, return a double containing the value of n taken to the falling factorial power of k.

### Definition

 Class: FallingFactorialPower Method: compute Parameters: int, int Returns: double Method signature: double compute(int n, int k) (be sure your method is public)

### Notes

-Your return must have relative or absolute error less than 1E-9.

### Constraints

-n will be between 1 and 10, inclusive.
-k will be between -5 and 5, inclusive.

### Examples

0)

 `7` `3`
`Returns: 210.0`
 7^^3=7*6*5=210.
1)

 `10` `1`
`Returns: 10.0`
2)

 `5` `0`
`Returns: 1.0`
3)

 `3` `-1`
`Returns: 0.25`
4)

 `2` `-3`
`Returns: 0.016666666666666666`

#### Problem url:

http://www.topcoder.com/stat?c=problem_statement&pm=9761

#### Problem stats url:

http://www.topcoder.com/tc?module=ProblemDetail&rd=12177&pm=9761

Petr

#### Testers:

PabloGilberto , Olexiy , Jan_Kuipers , ivan_metelsky

Simple Math