<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>

import math 
from collections import defaultdict
class Solution:
    def countAnagrams(self, s: str) -> int:
        MOD = 10**9 + 7
        MAX = max(len(word) for word in s.split(' '))
        
        factorial = [1] * (MAX + 1)
        for i in range(1, MAX + 1):
            factorial[i] = (factorial[i - 1] * i) % MOD

        inv_factorial = [1] * (MAX + 1)
        inv_factorial[MAX] = pow(factorial[MAX], MOD - 2, MOD)
        for i in range(MAX - 1, -1, -1):
            inv_factorial[i] = (inv_factorial[i + 1] * (i + 1)) % MOD

        def countPerms(w):
            cs = defaultdict(int)
            for c in w:
                cs[c] += 1
            d = 1
            for c in cs:
                d = (d * inv_factorial[cs[c]]) % MOD
            return (factorial[len(w)] * d) % MOD
                

        words = s.split(' ')
        result = 1
        for word in words:
            result = (result * countPerms(word)) % MOD 

        return result
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