<p>You are given a string <code>s</code> (<strong>0-indexed</strong>)​​​​​​. You are asked to perform the following operation on <code>s</code>​​​​​​ until you get a sorted string:</p>

<ol>
	<li>Find <strong>the largest index</strong> <code>i</code> such that <code>1 &lt;= i &lt; s.length</code> and <code>s[i] &lt; s[i - 1]</code>.</li>
	<li>Find <strong>the largest index</strong> <code>j</code> such that <code>i &lt;= j &lt; s.length</code> and <code>s[k] &lt; s[i - 1]</code> for all the possible values of <code>k</code> in the range <code>[i, j]</code> inclusive.</li>
	<li>Swap the two characters at indices <code>i - 1</code>​​​​ and <code>j</code>​​​​​.</li>
	<li>Reverse the suffix starting at index <code>i</code>​​​​​​.</li>
</ol>

<p>Return <em>the number of operations needed to make the string sorted.</em> Since the answer can be too 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;cba&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> The simulation goes as follows:
Operation 1: i=2, j=2. Swap s[1] and s[2] to get s=&quot;cab&quot;, then reverse the suffix starting at 2. Now, s=&quot;cab&quot;.
Operation 2: i=1, j=2. Swap s[0] and s[2] to get s=&quot;bac&quot;, then reverse the suffix starting at 1. Now, s=&quot;bca&quot;.
Operation 3: i=2, j=2. Swap s[1] and s[2] to get s=&quot;bac&quot;, then reverse the suffix starting at 2. Now, s=&quot;bac&quot;.
Operation 4: i=1, j=1. Swap s[0] and s[1] to get s=&quot;abc&quot;, then reverse the suffix starting at 1. Now, s=&quot;acb&quot;.
Operation 5: i=2, j=2. Swap s[1] and s[2] to get s=&quot;abc&quot;, then reverse the suffix starting at 2. Now, s=&quot;abc&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aabaa&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The simulation goes as follows:
Operation 1: i=3, j=4. Swap s[2] and s[4] to get s=&quot;aaaab&quot;, then reverse the substring starting at 3. Now, s=&quot;aaaba&quot;.
Operation 2: i=4, j=4. Swap s[3] and s[4] to get s=&quot;aaaab&quot;, then reverse the substring starting at 4. Now, s=&quot;aaaab&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 3000</code></li>
	<li><code>s</code>​​​​​​ consists only of lowercase English letters.</li>
</ul>

class Solution:
    @lru_cache(None)
    def f(self,i):
        return 1 if i <= 1 else (self.f(i-1)*i) % (10**9+7)

    def makeStringSorted(self, s):
        N, n, out = 10**9 + 7, len(s), 0
        cnt = [0]*26
        for i in range(n-1, -1, -1):
            ind = ord(s[i]) - ord("a")
            cnt[ind] += 1
            ans = sum(cnt[:ind]) * self.f(n-i-1)
            for j in range(26):
                ans = ans * pow(self.f(cnt[j]), N-2, N) % N
            out += ans
        return out % N
28000