You are given two strings `s` and `t`. In one step, you can append **any character** to either `s` or `t`.

Return _the minimum number of steps to make_ `s` _and_ `t` _**anagrams** of each other._

An **anagram** of a string is a string that contains the same characters with a different (or the same) ordering.

**Example 1:**

**Input:** s =  "**lee**tco**de** ", t =  "co**a**t**s** "
**Output:** 7
**Explanation:** 
- In 2 steps, we can append the letters in  "as " onto s =  "leetcode ", forming s =  "leetcode**as** ".
- In 5 steps, we can append the letters in  "leede " onto t =  "coats ", forming t =  "coats**leede** ".
 "leetcodeas " and  "coatsleede " are now anagrams of each other.
We used a total of 2 + 5 = 7 steps.
It can be shown that there is no way to make them anagrams of each other with less than 7 steps.

**Example 2:**

**Input:** s =  "night ", t =  "thing "
**Output:** 0
**Explanation:** The given strings are already anagrams of each other. Thus, we do not need any further steps.

**Constraints:**

*   `1 <= s.length, t.length <= 2 * 105`
*   `s` and `t` consist of lowercase English letters.


def count_vowel_substrings(word: str) -> int:
    count = 0
    n = len(word)
    vowels = set()

    for i in range(n):
        if word[i] in "aeiou":
            vowels.clear()
            for j in range(i, n):
                if word[j] in "aeiou":
                    vowels.add(word[j])
                if len(vowels) == 5:
                    count += 1
    return count
