<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>

class Solution:
    def countAnagrams(self, s: str) -> int:
        MOD = 10**9 + 7
        
        # Precompute factorials and their inverses
        max_factorial = 100000
        factorial = [1] * (max_factorial + 1)
        for i in range(2, max_factorial + 1):
            factorial[i] = factorial[i - 1] * i % MOD
        
        # Function to compute modular inverse using Fermat's Little Theorem
        def mod_inverse(x, mod):
            return pow(x, mod - 2, mod)
        
        # Precompute inverse factorials
        inverse_factorial = [1] * (max_factorial + 1)
        for i in range(2, max_factorial + 1):
            inverse_factorial[i] = mod_inverse(factorial[i], MOD)
        
        # Split the string into words
        words = s.split()
        
        # Calculate the number of anagrams for each word
        result = 1
        for word in words:
            # Count frequency of each character
            freq = {}
            for char in word:
                if char in freq:
                    freq[char] += 1
                else:
                    freq[char] = 1
            
            # Calculate the number of anagrams for this word
            word_length = len(word)
            word_anagrams = factorial[word_length]
            for count in freq.values():
                word_anagrams = word_anagrams * inverse_factorial[count] % MOD
            
            # Multiply the result with the number of anagrams for this word
            result = result * word_anagrams % MOD
        
        return result

# Example usage:
# solution = Solution()
# print(solution.countAnagrams("too hot"))  # Output: 18
# print(solution.countAnagrams("aa"))       # Output: 1
25400