Complete the function that counts the number of unique consonants in a string (made up of printable ascii characters).

Consonants are letters used in English other than `"a", "e", "i", "o", "u"`. We will count `"y"` as a consonant.

Remember, your function needs to return the number of unique consonants - disregarding duplicates. For example, if the string passed into the function reads `"add"`, the function should return `1` rather than `2`, since `"d"` is a duplicate.

Similarly, the function should also disregard duplicate consonants of differing cases. For example, `"Dad"` passed into the function should return `1` as `"d"` and `"D"` are duplicates.

## Examples

```
"add" ==> 1
"Dad" ==> 1
"aeiou" ==> 0
"sillystring" ==> 7
"abcdefghijklmnopqrstuvwxyz" ==> 21
"Count my unique consonants!!" ==> 7
```

Difficulty: introductory

Starter code:

def count_consonants(text):
	

Solutions:
CONSONANTS = set('bcdfghjklmnpqrstvwxyz')

def count_consonants(text):
    return len(CONSONANTS.intersection(text.lower()))

def count_consonants(text):
    return len(set(filter(str.isalpha,text.lower())) - set("aeiou"))

def count_consonants(s):
    return len({c for c in s.lower() if c in 'bcdfghjklmnpqrstvwxyz'})

count_consonants=lambda s: len(set(l for l in s.lower() if l in "bcdfghjklmnpqrstvwxyz"))

CONSONANTS = set("bcdfghjklmnpqrstvwxyz")

def count_consonants(text):
    return len(CONSONANTS & set(text.lower()))

def count_consonants(text):
    return len({ch for ch in text.lower() if ch in 'bcdfghjklmnpqrstvwxyz'})

def count_consonants(text):
    return len(set(c for c in text.lower() if c in "bcdfghjklmnpqrstvwxxyz"))


def count_consonants(text):
    return len(set(list(text.lower())).intersection(set(list('bcdfghjklmnpqrstvwxyz'))))

def count_consonants(text):
    # Your code here:
    vowels = ['a', 'e', 'i', 'o', 'u']
    consonants = []
    for letter in text:
        letter = letter.lower()
        conditions = [letter not in vowels,
                      letter not in consonants,
                      letter.isalpha()]
        if all(conditions):
            consonants.append(letter)
    return len(consonants)

def count_consonants(s):
    return len(set(c for c in s.lower() if c in "bcdfghjklmnpqrstvwxyz"))