<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>

MOD = 10**9 + 7

class Combinatorics:
    def __init__(self, pre_compute_limit=0):
        self.fact = [1] * pre_compute_limit
        self.invfact = [1] * pre_compute_limit
        for i in range(1, pre_compute_limit):
            self.fact[i] = (self.fact[i - 1] * i) % MOD
            self.invfact[i] = (self.invfact[i - 1] * self.mod_inverse(i)) % MOD

    def mod_inverse(self, a):
        return pow(a, MOD - 2, MOD)

    def ncr(self, n, r):
        if n < r:
            return 0
        return (self.fact[n] * self.invfact[r] * self.invfact[n - r]) % MOD

    def npr(self, n, r):
        if n < r:
            return 0
        return (self.fact[n] * self.invfact[n - r]) % MOD

    def ncr_without_memo(self, n, r):
        if n < r:
            return 0
        num = den = 1
        for i in range(r):
            num = (num * (n - i)) % MOD
            den = (den * (i + 1)) % MOD
        return (num * pow(den, MOD - 2, MOD)) % MOD

    def npr_without_memo(self, n, r):
        if n < r:
            return 0
        num = 1
        for i in range(n, n - r, -1):
            num = (num * i) % MOD
        return num

class Solution:
    def countAnagrams(self, s: str) -> int:
        n = len(s)
        com = Combinatorics(n+5)

        words = s.split()

        ans = 1

        for word in words:
            c = Counter(word)
            ans*=com.fact[len(word)]
            for v in c.values():
                ans*=com.invfact[v]
            ans%=MOD
        
        return ans

951