Given a string s, return the last substring of s in lexicographical order.
 
Example 1:
Input: "abab"
Output: "bab"
Explanation: The substrings are ["a", "ab", "aba", "abab", "b", "ba", "bab"]. The lexicographically maximum substring is "bab".

Example 2:
Input: "leetcode"
Output: "tcode"

 
Note:

1 <= s.length <= 4 * 10^5
s contains 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    def lastSubstring(self, s: str) -> str:\n        # 因为字典顺序，可以看出 ab abc 两个字符中， abc 更大，因为两个字符相同时，谁更长谁order 更大\n        # 所以，可以确定我们只需要找到 最大substring 的起始点，然后把前面都切掉，后缀就是答案\n        # 利用双指针来找起始点\n        # i j 两个指针 指向数组的两个字符，维持 i +k ， j +k 两个substring\n        # 每次比较，如果i 的更大，那么j就要更新到j + k + 1， 因为在j ～ j + k 之间的字符都是小于i的，不然也不会等到现在更新\n        # 如果 j更大，那么用j的去更新i，j移动到自己下一个位置 找最大string，继续去比较\n        i = 0\n        j = 1\n        k = 0\n        while j + k < len(s):\n            if s[i + k] == s[j + k]:\n                k += 1\n                continue\n            elif s[i + k] > s[j + k]:\n                j = j + k + 1\n            else:\n                i = j\n                j = i + 1\n            k = 0\n        return s[i:]\n        \n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'lastSubstring'
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: 5579