Given an array, return the difference between the count of even numbers and the count of odd numbers. `0` will be considered an even number. 

```
For example:
solve([0,1,2,3]) = 0 because there are two even numbers and two odd numbers. Even - Odd = 2 - 2 = 0.  
```

Let's now add two letters to the last example: 
```
solve([0,1,2,3,'a','b']) = 0. Again, Even - Odd = 2 - 2 = 0. Ignore letters. 
```

The input will be an array of lowercase letters and numbers only. 

In some languages (Haskell, C++, and others), input will be an array of strings:

```
solve ["0","1","2","3","a","b"] = 0 
```

Good luck!

If you like this Kata, please try: 

[Longest vowel chain](https://www.codewars.com/kata/59c5f4e9d751df43cf000035)

[Word values](https://www.codewars.com/kata/598d91785d4ce3ec4f000018)

Difficulty: EASY
Starter code: def solve(a):
	

Solutions:
def solve(a):
	return sum((1 if v % 2 == 0 else -1 for v in a if type(v) == int))


def solve(a):
	return sum((isinstance(x, int) and [1, -1][x % 2] for x in a))


def solve(a):
	diff = 0
	for e in a:
		if type(e) == int:
			if e % 2:
				diff -= 1
			else:
				diff += 1
	return diff


def solve(a):
	lenList = len(a)
	evenEntries = [a[i] for i in range(0, lenList, 1) if str(a[i]).isdigit() and a[i] % 2 == 0]
	oddEntries = [a[i] for i in range(0, lenList, 1) if str(a[i]).isdigit() and a[i] % 2 == 1]
	amtEvenEntries = len(evenEntries)
	amtOddEntries = len(oddEntries)
	return amtEvenEntries - amtOddEntries


def solve(a):
	return sum(((-1) ** x for x in a if isinstance(x, int)))


def solve(a):
	return len([i for i in a if type(i) == int and (not i % 2)]) - len([i for i in a if type(i) == int and i % 2])


def solve(a):
	a = [i for i in a if isinstance(i, int)]
	return sum((1 for i in a if i % 2 == 0)) - sum((1 for i in a if i % 2))