Given a string `word`, return _the **sum of the number of vowels** (_`'a'`, `'e'`_,_ `'i'`_,_ `'o'`_, and_ `'u'`_)_ _in every substring of_ `word`.

A **substring** is a contiguous (non-empty) sequence of characters within a string.

**Note:** Due to the large constraints, the answer may not fit in a signed 32-bit integer. Please be careful during the calculations.

**Example 1:**

**Input:** word =  "aba "
**Output:** 6
**Explanation:** 
All possible substrings are:  "a ",  "ab ",  "aba ",  "b ",  "ba ", and  "a ".
-  "b " has 0 vowels in it
-  "a ",  "ab ",  "ba ", and  "a " have 1 vowel each
-  "aba " has 2 vowels in it
Hence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6. 

**Example 2:**

**Input:** word =  "abc "
**Output:** 3
**Explanation:** 
All possible substrings are:  "a ",  "ab ",  "abc ",  "b ",  "bc ", and  "c ".
-  "a ",  "ab ", and  "abc " have 1 vowel each
-  "b ",  "bc ", and  "c " have 0 vowels each
Hence, the total sum of vowels = 1 + 1 + 1 + 0 + 0 + 0 = 3.

**Example 3:**

**Input:** word =  "ltcd "
**Output:** 0
**Explanation:** There are no vowels in any substring of  "ltcd ".

**Constraints:**

*   `1 <= word.length <= 105`
*   `word` consists of lowercase English letters.


def count_vowels(word):
    sum = 0
    count = 0
    for letter in word:
        if letter in "aeiou":
            count += 1
        sum += count
    return sum


def count_vowels_in_substrings(word):
    return count_vowels(word)
