Return the number of distinct non-empty substrings of text that can be written as the concatenation of some string with itself (i.e. it can be written as a + a where a is some string).
 
Example 1:
Input: text = "abcabcabc"
Output: 3
Explanation: The 3 substrings are "abcabc", "bcabca" and "cabcab".

Example 2:
Input: text = "leetcodeleetcode"
Output: 2
Explanation: The 2 substrings are "ee" and "leetcodeleetcode".

 
Constraints:

1 <= text.length <= 2000
text has only lowercase English letters.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

def _disabled_input(*args, **kwargs):
    raise RuntimeError("input() is disabled in callable mode")
builtins.input = _disabled_input

ns = {"__name__": "__judge__"}  # prevent if __name__ == "__main__" blocks
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "class Solution:\n    \n    def distinctEchoSubstrings2(self, text):\n        base = 29\n        mod = 10**9 + 7\n        \n        aset = set()\n        n = len(text)\n        \n        thash = [0] * (n+1) # thash[i] is the hash value for text[0..i-1] , 从text[0]到text[i-1]都是包括的！\n        tpow = [0] * (n+1) # tpow[i] = base ^ i\n        tpow[0] = 1\n        \n        def getHash(l, r):\n            return (thash[r] - thash[l]*tpow[r-l] % mod + mod) % mod # 快速获取一个区间的hash! 这个理解起来有难度！\n            #       全体0..r    0..l, 幂的差r-l\n        \n        for i in range(1, n+1):\n            #thash[i] = (thash[i-1] * base + ord(text[i-1])-ord('a')) % mod # something is wrong if we use ord(text[i-1]) - ord('a')\n            thash[i] = (thash[i-1] * base + ord(text[i-1])) % mod # 简单的进位叠加，这个还算容易记住！\n            tpow[i] = tpow[i-1] * base % mod # 简单的进位叠加！还算容易记住！\n        \n        for alen in range(1, n//2+1): # abab case? 而且代码并不直观的感觉！\n            l = 0\n            r = alen\n            counter = 0\n            while l < n - alen:\n                if text[l] == text[r]:\n                    counter += 1\n                else:\n                    counter = 0\n                \n                if counter == alen:\n                    aset.add(getHash(l-alen+1, l+1))\n                    counter -= 1\n                l += 1\n                r += 1\n        return len(aset)\n        '''public int distinctEchoSubstrings(String text) {\n\\t\\tSet<Long> set = new HashSet<>();\n\\t\\tint n = text.length();\n\\t\\tpreprocess(n, text);\n\n\\t\\tfor (int len = 1; len <= n / 2; len++) {\n\\t\\t\\tfor (int l = 0, r = len, counter = 0; l < n - len; l++, r++) {\n\\t\\t\\t\\tif (text.charAt(l) == text.charAt(r)) {\n\\t\\t\\t\\t\\tcounter++;\n\\t\\t\\t\\t} else {\n\\t\\t\\t\\t\\tcounter = 0;\n\\t\\t\\t\\t}\n                \n                // 这个操作可以加速？！并不是所有的都是直接call getHash的！\n\\t\\t\\t\\tif (counter == len) {\n\\t\\t\\t\\t\\tset.add(getHash(l - len + 1, l + 1));\n\\t\\t\\t\\t\\tcounter--;\n\\t\\t\\t\\t}\n\\t\\t\\t}\n\\t\\t}\n\n\\t\\treturn set.size();\n\\t}'''\n        \n    def distinctEchoSubstrings(self, text):\n        base = 29\n        mod = 10**9 + 7\n        aset = set()\n        n = len(text)\n        \n        thash = [0] * (n+1)\n        tpow = [0] * (n+1)\n        tpow[0] = 1\n        \n        def getHash(l, r):\n            return (thash[r] - thash[l]*tpow[r-l] % mod + mod) % mod\n        for i in range(1, n+1):\n            thash[i] = (thash[i-1] * base + ord(text[i-1])) % mod\n            tpow[i] = tpow[i-1]*base % mod\n        \n        for i in range(n):\n            for alen in range(2, n-i+1, 2): # i=n-1 then alen=1 is okay, i=0 then alen=n is okay; which means that i+alen=n! so alen = n - i is the upper bound\n                mid = i + alen//2\n                hash1 = getHash(i, mid)\n                hash2 = getHash(mid, i+alen)\n                if hash1 == hash2:\n                    aset.add(hash1)\n        return len(aset)\n                \n\n    \n    def distinctEchoSubstrings1(self, text: str) -> int:\n        base = 29\n        mod = 10**9 + 7\n        \n        aset = set()\n        n = len(text)\n        \n        thash = [0] * (n+1) # thash[i] is the hash value for text[0..i-1] , 从text[0]到text[i-1]都是包括的！\n        tpow = [0] * (n+1) # tpow[i] = base ^ i\n        tpow[0] = 1\n        \n        def getHash(l, r):\n            return (thash[r] - thash[l]*tpow[r-l] % mod + mod) % mod # 快速获取一个区间的hash! 这个理解起来有难度！\n            #       全体0..r    0..l, 幂的差r-l\n        \n        for i in range(1, n+1):\n            #thash[i] = (thash[i-1] * base + ord(text[i-1])-ord('a')) % mod # something is wrong if we use ord(text[i-1]) - ord('a')\n            thash[i] = (thash[i-1] * base + ord(text[i-1])) % mod # 简单的进位叠加，这个还算容易记住！\n            tpow[i] = tpow[i-1] * base % mod # 简单的进位叠加！还算容易记住！\n            \n        # 理解一下：\n        # abcd -> \n        # thash[0] = 0\n        # thash[1] = (0*29 + ord('a')) % mod = ord('a') % mod\n        # thash[2] = [ ord('a')*29 + ord('b') ] % mod\n        # thash[3] = [ ord('a')*29*29 + ord('b') * 29 + ord('c')] % mod\n        # thash[4] = [ ord('a')*29*29*29 + ord('b') * 29*29 + ord('c')*29 + ord('d')] % mod\n        \n        # tpow[1] = 1*29 % mod = 29 % mod\n        # tpow[2] = 29*29 % mod\n        # tpow[3] = 29*29*29 % mod\n        # tpow[4] = 29*29*29*29 % mod\n        \n        # abab 的时候\n        # thash[1] = [ ord('a') ] % mod\n        # thash[2] = [ ord('a')*29 + ord('b') ] % mod\n        # thash[3] = [ ord('a')*29*29 + ord('b') * 29 + ord('a')] % mod\n        # thash[4] = [ ord('a')*29*29*29 + ord('b') * 29*29 + ord('a')*29 + ord('b')] % mod\n        # tpow[4] = 29*29*29*29 % mod\n        \n        # i=0, alen=4 -> abab\n        # mid=2\n        # so, it is: \n        # getHash(0, 2) -> (ord('a')*29 + ord('b') - 0*tpow[2] + mod) % mod \n        #               -> (ord('a')*29 + ord('b') + mod) % mod\n        # getHash(2, 4) -> (ord('a')*29*29*29 + ord('b') * 29*29 + ord('a')*29 + ord('b') - (ord('a')*29 + ord('b'))*29*29 + mod) % mod\n        #               -> (ord('a')*29 + ord('b') + mod) % mod\n        \n        \n        for i in range(n):\n            for alen in range(2, n-i+1, 2): # 遍历所有长度为偶数的子串（中间切开，然后比较左右）\n                mid = i + alen//2\n                hash1 = getHash(i, mid)\n                hash2 = getHash(mid, i+alen)\n                if hash1 == hash2: # problem, even hash1 == hash2, not necessary mean that text[i,mid] = text[mid, i+alen]!\n                    aset.add(hash1)\n        return len(aset)\n        \n        \n        '''\n        class Solution {\n        long BASE = 29L, MOD = 1000000007L;\n        public int distinctEchoSubstrings(String str) {\n            HashSet<Long> set = new HashSet<>();\n            int n = str.length();\n            long[] hash = new long[n + 1]; // hash[i] is hash value from str[0..i]\n            long[] pow = new long[n + 1]; // pow[i] = BASE^i\n            pow[0] = 1;\n            for (int i = 1; i <= n; i++) {\n                hash[i] = (hash[i - 1] * BASE + str.charAt(i - 1)) % MOD;\n                pow[i] = pow[i - 1] * BASE % MOD;\n            }\n            for (int i = 0; i < n; i++) {\n                for (int len = 2; i + len <= n; len += 2) {\n                    int mid = i + len / 2;\n                    long hash1 = getHash(i, mid, hash, pow);\n                    long hash2 = getHash(mid, i + len, hash, pow);\n                    if (hash1 == hash2) set.add(hash1);\n                }\n            }\n            return set.size();\n        }\n\n        long getHash(int l, int r, long[] hash, long[] pow) {\n            return (hash[r] - hash[l] * pow[r - l] % MOD + MOD) % MOD;\n        }\n    }\n    '''"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'distinctEchoSubstrings'
if _fn_name:
    if "Solution" in ns:
        try:
            _obj = ns["Solution"]()
            if hasattr(_obj, _fn_name):
                _target = getattr(_obj, _fn_name)
        except Exception:
            _target = None
    if _target is None and _fn_name in ns and callable(ns[_fn_name]):
        _target = ns[_fn_name]
# Fallback: if there's exactly one user-defined function, use it
if _target is None:
    _cands = [v for k, v in ns.items() if callable(v) and getattr(v, "__module__", "") in ("__judge__", "__main__")]
    _user_funcs = [c for c in _cands if getattr(c, "__name__", "").startswith("_") is False]
    if len(_user_funcs) == 1:
        _target = _user_funcs[0]
if _target is None:
    raise RuntimeError(f"Could not resolve callable '{_fn_name}'")

_raw = sys.stdin.read().strip()
if _raw:
    _parsed = json.loads(_raw)
    if isinstance(_parsed, dict):
        _kwargs = _parsed.get("kwargs", {})
        _args = _parsed.get("args", [])
    else:
        _args, _kwargs = _parsed, {}
else:
    _args, _kwargs = [], {}
if not isinstance(_args, (list, tuple)):
    _args = [_args]
_res = _target(*_args, **_kwargs)
try:
    sys.stdout.write(repr(_res) + "\n")
except Exception:
    pass

```

Measured memory_bytes: 7482