<p>You are given a string <code>s</code> and a positive integer <code>k</code>.</p>

<p>Let <code>vowels</code> and <code>consonants</code> be the number of vowels and consonants in a string.</p>

<p>A string is <strong>beautiful</strong> if:</p>

<ul>
	<li><code>vowels == consonants</code>.</li>
	<li><code>(vowels * consonants) % k == 0</code>, in other terms the multiplication of <code>vowels</code> and <code>consonants</code> is divisible by <code>k</code>.</li>
</ul>

<p>Return <em>the number of <strong>non-empty beautiful substrings</strong> in the given string</em> <code>s</code>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>

<p><strong>Vowel letters</strong> in English are <code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, and <code>&#39;u&#39;</code>.</p>

<p><strong>Consonant letters</strong> in English are every letter except vowels.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;baeyh&quot;, k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 beautiful substrings in the given string.
- Substring &quot;b<u>aeyh</u>&quot;, vowels = 2 ([&quot;a&quot;,e&quot;]), consonants = 2 ([&quot;y&quot;,&quot;h&quot;]).
You can see that string &quot;aeyh&quot; is beautiful as vowels == consonants and vowels * consonants % k == 0.
- Substring &quot;<u>baey</u>h&quot;, vowels = 2 ([&quot;a&quot;,e&quot;]), consonants = 2 ([&quot;b&quot;,&quot;y&quot;]).
You can see that string &quot;baey&quot; is beautiful as vowels == consonants and vowels * consonants % k == 0.
It can be shown that there are only 2 beautiful substrings in the given string.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abba&quot;, k = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are 3 beautiful substrings in the given string.
- Substring &quot;<u>ab</u>ba&quot;, vowels = 1 ([&quot;a&quot;]), consonants = 1 ([&quot;b&quot;]).
- Substring &quot;ab<u>ba</u>&quot;, vowels = 1 ([&quot;a&quot;]), consonants = 1 ([&quot;b&quot;]).
- Substring &quot;<u>abba</u>&quot;, vowels = 2 ([&quot;a&quot;,&quot;a&quot;]), consonants = 2 ([&quot;b&quot;,&quot;b&quot;]).
It can be shown that there are only 3 beautiful substrings in the given string.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bcdf&quot;, k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no beautiful substrings in the given string.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= 1000</code></li>
	<li><code>s</code> consists of only English lowercase letters.</li>
</ul>

k0 = [i for i in range(1001)]
for a in reversed(range(2, len(k0))):
    a2 = a*a
    for i in range(a2, len(k0), a2):
        if k0[i]==i:
            k0[i] //= a

vowels = "aeiou"

class Solution:
    def beautifulSubstrings(self, s: str, k: int) -> int:
        counter = {0: {0: 1}}
        total = 0
        vowel_cnt = 0
        k = k0[k]
        
        for i, ch in enumerate(s, 1):
            if ch in vowels:
                vowel_cnt += 1
            diff = i-2*vowel_cnt
            idx = vowel_cnt % k
            if diff in counter:
                d = counter[diff]
                total += d.get(idx % k, 0)
                if idx not in d:
                    d[idx] = 1
                else:
                    d[idx] += 1
            else:
                counter[diff] = {idx: 1} 
            
        return total
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