<p>You are given a string <code>s</code> containing one or more words. Every consecutive pair of words is separated by a single space <code>&#39; &#39;</code>.</p>

<p>A string <code>t</code> is an <strong>anagram</strong> of string <code>s</code> if the <code>i<sup>th</sup></code> word of <code>t</code> is a <strong>permutation</strong> of the <code>i<sup>th</sup></code> word of <code>s</code>.</p>

<ul>
	<li>For example, <code>&quot;acb dfe&quot;</code> is an anagram of <code>&quot;abc def&quot;</code>, but <code>&quot;def cab&quot;</code>&nbsp;and <code>&quot;adc bef&quot;</code> are not.</li>
</ul>

<p>Return <em>the number of <strong>distinct anagrams</strong> of </em><code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;too hot&quot;
<strong>Output:</strong> 18
<strong>Explanation:</strong> Some of the anagrams of the given string are &quot;too hot&quot;, &quot;oot hot&quot;, &quot;oto toh&quot;, &quot;too toh&quot;, and &quot;too oht&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aa&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is only one anagram possible for the given string.</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters and spaces <code>&#39; &#39;</code>.</li>
	<li>There is single space between consecutive words.</li>
</ul>

from collections import Counter

MOD = 10**9 + 7

def precompute_factorials_and_inverses(n: int):
    factorials = [1] * (n + 1)
    inverses = [1] * (n + 1)
    
    for i in range(2, n + 1):
        factorials[i] = factorials[i - 1] * i % MOD
    
    inverses[n] = pow(factorials[n], MOD - 2, MOD)
    for i in range(n - 1, 0, -1):
        inverses[i] = inverses[i + 1] * (i + 1) % MOD
    
    return factorials, inverses

def countAnagrams(s: str) -> int:
    words = s.split()
    max_len = max(len(word) for word in words)
    factorials, inverses = precompute_factorials_and_inverses(max_len)
    
    def count_word_anagrams(word):
        counter = Counter(word)
        n = len(word)
        result = factorials[n]
        for count in counter.values():
            result = result * inverses[count] % MOD
        return result
    
    total_anagrams = 1
    for word in words:
        total_anagrams = total_anagrams * count_word_anagrams(word) % MOD
    
    return total_anagrams

# Example usage
print(countAnagrams("too hot"))  # Should print 18
print(countAnagrams("aa"))       # Should print 1



class Solution:
    def countAnagrams(self, *args, **kwargs):
        return countAnagrams(*args, **kwargs)


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