<p>Return the number of <strong>distinct</strong> non-empty substrings of <code>text</code>&nbsp;that can be written as the concatenation of some string with itself (i.e. it can be written as <code>a + a</code>&nbsp;where <code>a</code> is some string).</p>

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

<pre>
<strong>Input:</strong> text = &quot;abcabcabc&quot;
<strong>Output:</strong> 3
<b>Explanation: </b>The 3 substrings are &quot;abcabc&quot;, &quot;bcabca&quot; and &quot;cabcab&quot;.
</pre>

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

<pre>
<strong>Input:</strong> text = &quot;leetcodeleetcode&quot;
<strong>Output:</strong> 2
<b>Explanation: </b>The 2 substrings are &quot;ee&quot; and &quot;leetcodeleetcode&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= text.length &lt;= 2000</code></li>
	<li><code>text</code>&nbsp;has only lowercase English letters.</li>
</ul>

class Solution:
    def distinctEchoSubstrings(self, text: str) -> int:
        n = len(text)
        M = 10 ** 10 + 7
        x = 26
        p = [1]

        for i in range(1, n):
            p.append((p[-1] * x) % M)

        res = set()

        for length in range(1, n//2+1):
            hashed = [-1] * n
            curr = 0
            
            for j in range(length):
                v = ord(text[j])-ord('a')+1
                curr = (curr + v * p[length-j-1]) % M
            hashed[length-1] = curr

            for j in range(length, n):
                v = ord(text[j])-ord('a')+1
                v2 = ord(text[j-length])-ord('a')+1
                curr = (curr - v2 * p[length-1] + M) % M
                curr = (curr * x) % M
                curr = (curr + v) % M
                hashed[j] = curr
                if hashed[j] == hashed[j-length]:
                    res.add(hashed[j])
                
        return len(res)
16900