<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
        fact = [1]
        for i in range(1, 10**5 + 1):
            fact.append(fact[-1] * i % MOD)
        
        def mod_inverse(a, p):
            def extended_gcd(a, b):
                x0, x1, y0, y1 = 1, 0, 0, 1
                while b != 0:
                    q, a, b = a // b, b, a % b
                    x0, x1 = x1, x0 - q * x1
                    y0, y1 = y1, y0 - q * y1
                return a, x0, y0
            
            g, x, y = extended_gcd(a, p)
            return x % p
        
        def count_anagrams(word):
            n = len(word)
            freq = [0] * 26
            for c in word:
                freq[ord(c) - ord('a')] += 1
            ans = fact[n]
            for f in freq:
                ans = ans * mod_inverse(fact[f], MOD) % MOD
            return ans
        
        ans = 1
        for word in s.split():
            ans = ans * count_anagrams(word) % MOD
        return ans
663