<p>You are given a <strong>0-indexed</strong> string <code>s</code>, a string <code>a</code>, a string <code>b</code>, and an integer <code>k</code>.</p>

<p>An index <code>i</code> is <strong>beautiful</strong> if:</p>

<ul>
	<li><code>0 &lt;= i &lt;= s.length - a.length</code></li>
	<li><code>s[i..(i + a.length - 1)] == a</code></li>
	<li>There exists an index <code>j</code> such that:
	<ul>
		<li><code>0 &lt;= j &lt;= s.length - b.length</code></li>
		<li><code>s[j..(j + b.length - 1)] == b</code></li>
		<li><code>|j - i| &lt;= k</code></li>
	</ul>
	</li>
</ul>

<p>Return <em>the array that contains beautiful indices in <strong>sorted order from smallest to largest</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;isawsquirrelnearmysquirrelhouseohmy&quot;, a = &quot;my&quot;, b = &quot;squirrel&quot;, k = 15
<strong>Output:</strong> [16,33]
<strong>Explanation:</strong> There are 2 beautiful indices: [16,33].
- The index 16 is beautiful as s[16..17] == &quot;my&quot; and there exists an index 4 with s[4..11] == &quot;squirrel&quot; and |16 - 4| &lt;= 15.
- The index 33 is beautiful as s[33..34] == &quot;my&quot; and there exists an index 18 with s[18..25] == &quot;squirrel&quot; and |33 - 18| &lt;= 15.
Thus we return [16,33] as the result.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, a = &quot;a&quot;, b = &quot;a&quot;, k = 4
<strong>Output:</strong> [0]
<strong>Explanation:</strong> There is 1 beautiful index: [0].
- The index 0 is beautiful as s[0..0] == &quot;a&quot; and there exists an index 0 with s[0..0] == &quot;a&quot; and |0 - 0| &lt;= 4.
Thus we return [0] as the result.
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= a.length, b.length &lt;= 10</code></li>
	<li><code>s</code>, <code>a</code>, and <code>b</code> contain only lowercase English letters.</li>
</ul>

class RollingHash:
    def __init__(self ,s):
        s = [ord(char) - 97 for char in s]
        self.mod = mod = 344555666677777
        self.basePower = basePower = 10 ** 7 + 7
        #fermat's little theorem since basePower and mod are co-prime
        self.invbase = invbase = pow(basePower, mod - 2, mod)
        self.prefix = prefix = [0] # r - l in query so for 1st prefix -> l needs to be 0 right?
        self.invprefix = invprefix = [1]
        hash, power, invpower = 0, 1, 1

        for x in s:
            hash = (hash + x * power) % mod
            power = (power * basePower) % mod
            invpower = (invpower * invbase) % mod
            prefix.append(hash)
            invprefix.append(invpower)
    
    def query(self, l, r):
        return ((self.prefix[r] - self.prefix[l]) * self.invprefix[l]) % self.mod


class Solution:
    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:
        S = RollingHash(s)
        A = RollingHash(a)
        B = RollingHash(b)
        l = 0
        ans = []
        for r in range(len(b), len(s) + 1):
            if S.query(l, r) == B.query(0, len(b)):
                ans.append(l)
            l += 1
        l = 0
        res = []
        for r in range(len(a), len(s) + 1):
            if S.query(l, r) == A.query(0, len(a)):
                pos = bisect_left(ans, l)
                if (pos < len(ans) and abs(ans[pos] - l) <= k) or (pos > 0 and abs(ans[pos - 1] - l) <= k):
                    res.append(l)
            l += 1
        return res
505