<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 functools import cache


def inv(b):
    m = 1_000_000_000+7
    r0, r1 = b, m
    s0, s1 = 1, 0

    while r1 > 0:
        qi = r0 // r1
        r1, r0 = r0 - qi * r1, r1
        s1, s0 = s0 - qi * s1, s1

    return s0
    
@cache
def fac(n):
    mod = 1_000_000_000+7
    
    if n == 0:
        return 1

    return (n * fac(n-1)) % mod

class Solution:
    def count(self, s: str):

        mod = 1_000_000_000+7

        freq = {}
        for c in s:
            freq[c] = freq.get(c, 0) + 1
        
        count = 1
        s = 0
        for k in freq:
            s += freq[k]
            count = count * math.comb(s, freq[k]) % mod

        return count
        # a = fac(len(s))
        # b = 1
        # for k in freq:
        #     b = (b*fac(freq[k])) % mod

        return (a * inv(b)) % mod

    def countAnagrams(self, s: str) -> int:
        words = s.split()

        total = 1
        mod = 1_000_000_000+7

        for word in words:
            total = (total * self.count(word)) % mod

        return total


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