<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 = 1000000007
        n = len(s)
        fact = [1]
        ifact = [1]
        for i in range(1, n+1):
            fact.append((fact[-1] * i)%mod)
            ifact.append(pow(fact[-1], -1, mod))
                
        def helper(arr):
            s = sum(arr)
            summ = fact[s]
            for i in arr:
                
                if i >=2:
                    summ*= ifact[i]
            # print(arr, s, resh)
            return (summ)%mod
                
                
        curr = [0]*26
        d = {}
        ans = 1
        cnt = 0
        res = 1
        for i in s:
            if i == ' ':
                if tuple(curr) in d:
                    d[tuple(curr)]  += 1
                else:
                    d[tuple(curr)]  = 1
                res*= helper(curr)
                # print(curr, 'inside i= " "', res)
                curr = [0]*26
            else:
                curr[ord(i) - ord('a')] += 1
        
        if curr != [0]*26:
            if tuple(curr) in d:
                d[tuple(curr)]  += 1
            else:
                d[tuple(curr)]  = 1
        # print(curr, 'last', res)
        res*= helper(curr)
        # divide = 1
        # for i in d.values():
        #     if i != 1:
        #         divide *= fact(i)
        # print(divide)
        # print(res%mod)
             
        return (res)%mod
        
            
            
                
                    
                    
                
        
        
611