<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

class Solution:
    def countAnagrams(self, s: str) -> int:
        # hot oht toh 
        # hot hto oht oth tho toh


        # h o o
        # o h o
        # o o h  non unique - len(string)

        # 
        ifact = [1]
        fact = [1]

        for i in range(1, len(s)+1):
            fact.append((fact[-1]*i) % (10 ** 9 + 7))
            ifact.append(pow(fact[-1], -1, 10 ** 9 + 7))
        
        sol = 1
        for sep in s.split(" "):
            c = Counter(sep)

            sol *= fact[sum(c.values())]
            sol %= (10 ** 9 + 7)

            # divide by every repeat 
            for k, v in c.items():
                sol *= ifact[v]
                sol %= (10 ** 9 + 7)

        return sol 
        # separated = s.split(" ")

        # sol = 1
        # for sep in separated:
        #     c = Counter(sep)
        #     mult = 1
        #     for i in range(1, len(sep)+1):
        #         mult *= i
        #         mult = mult % (10 ** 9 + 7)
            
        #     sol *= mult % (10 ** 9 + 7)
        #     sol = sol % (10 ** 9 + 7)
        #     for k, v in c.items():
        #         if v > 1:
        #             fact = 1
        #             for i in range(1, v+1):
        #                 fact *= i
        #                 fact = fact % (10 ** 9 + 7)
        #             sol /= fact % (10 ** 9 + 7)
        #             sol = sol % (10 ** 9 + 7)
            
        #     print(sep, mult, sol)
        # return int(sol) % ((10 ** 9) + 7)


        # h o x s
        # h x o s
        # h s x o
        # h s o x
        # h x s o
        # h o s x
559