Create a method (**JS**: function) `every` which returns every nth element of an array.

### Usage

With no arguments, `array.every` it returns every element of the array.  
With one argument, `array.every(interval)` it returns every `interval`th element.  
With two arguments, `array.every(interval, start_index)` it returns every `interval`th element starting at index `start_index` 

**Note**: due to fact JS translation ask for a function instead of an Array method the above uses will be :
```
javascript:                          ruby:
every(array)                         array.every
every(array, interval)               array.every(interval)
every(array, interval, start_index)  array.every(interval, start_index)

```

### Examples

### Notes
Test cases:  

`interval` will always be a positive integer (but may be larger than the size of the array).  
`start_index` will always be within the bounds of the array.  

Once you have completed this kata, try the **advanced version** (http://www.codewars.com/kata/every-nth-array-element-advanced) which allows negative intervals and unbounded start\_indexes 

### Translator notes
Ruby is the original language for this kata.

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Data structures

Starter code: def every(array, interval = 1, start_index = 0):
	

Solutions:
def every(lst, n=1, start=0):
	return lst[start::n]


def every(array, interval=None, start=None):
	return array[start::interval]


def every(arr, n=1, m=0):
	return arr[m::n]


def every(arr, n=1, s=0):
	return arr[s::n]


def every(arr, n=1, i=0):
	return arr[i::n]


def every(array, interval=None, start_index=None):
	return array[start_index::interval]


every = lambda arr, step=None, start=None: arr[start::step]


every = lambda s, n=1, f=0: s[f::n]


def every(n, x=1, y=0):
	return n[y::x]


def every(array, ii=0, si=0):
	return array[si:] if ii == 0 else array[si::ii]


def every(array, interval=1, start_index=0):
	li = []
	for i in range(start_index, len(array), interval):
		li.append(array[i])
	return li


def every(array, interval=1, index=0):
	return array[index::interval]


def every(array, interval=1, start_index=0):
	arr = []
	counter = 0
	for i in range(start_index, len(array)):
		if counter == interval:
			arr.append(array[i])
			counter = 0
		elif i == start_index:
			arr.append(array[start_index])
		counter += 1
	return arr


every = lambda lst, k=1, i=0: lst[i::k]


def every(array, interval=1, start_index=0):
	return array[start_index:][::interval]


def every(array=[0, 1, 2, 3, 4], interval=1, start_index=0):
	return array[start_index::interval]


def every(array, step=1, start=0):
	return array[start::step]


def every(array, interval=0, start_index=0):
	if interval == 0:
		return array
	return [x for (ind, x) in enumerate(array[start_index:]) if ind % interval == 0]


def every(arr, interval=1, start=0):
	flt = arr[start:]
	return list(flt[::interval])


def every(array, interval=1, start_index=0):
	res = []
	for i in range(start_index, len(array), interval):
		if i > len(array) + 1:
			break
		else:
			res.append(array[i])
	return res