Given a list of words, each word consists of English lowercase letters.
Let's say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2.  For example, "abc" is a predecessor of "abac".
A word chain is a sequence of words [word_1, word_2, ..., word_k] with k >= 1, where word_1 is a predecessor of word_2, word_2 is a predecessor of word_3, and so on.
Return the longest possible length of a word chain with words chosen from the given list of words.
 
Example 1:
Input: ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: one of the longest word chain is "a","ba","bda","bdca".

 
Note:

1 <= words.length <= 1000
1 <= words[i].length <= 16
words[i] only consists of English lowercase 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    \n    def isChild(self, parent, possible_child):\n        for i in range(len(possible_child)):\n            if possible_child[:i] + possible_child[i+1:] == parent:\n                return True \n        return False\n    \n    def longestStrChain(self, words: List[str]) -> int:\n        \n        # an intutitve idea is to create a graph of the given words, where every node is a word, and is a predecessor to all of its children. We can then simply DFS from root to node and find the longest path.\n        # the DFS would take O(n) where n is the length of words.\n        # Challenge: how to construct graph? We can loop once and bucket words by their length, and create a node per word, then starting from the smallest word length i, find all possible children of length i+1 and add them as edges do this until the end. \n        # bucketing and creating graph nodes would take O(n), connecting nodes would also take o(n) since we only check two buckets at a time linearly, and those buckets at most contain n nodes (and in that case other buckets are empty), so its O(n)\n        \n        \n        graph = {}\n        # word length buckets \n        word_lens = [[] for _ in range(17)]\n        # create nodes and bucket\n        for w in words:\n            graph[w] = []\n            word_lens[len(w)].append(w)\n        \n        # connect graph\n        for i in range(1,16):\n            parents = word_lens[i]\n            possible_children = word_lens[i+1]\n            for p in parents: \n                for c in possible_children:\n                    # check if c can possibly be a child of p\n                    # it is a possible child if \n                    if self.isChild(p,c):\n                        graph[p].append(c)\n       # do DFS \n        longest_len = 0\n        for node in graph.keys():\n            to_visit = [(node,1)]\n            \n            while(len(to_visit) != 0):\n                cur_node,depth = to_visit.pop()\n                longest_len = max(depth, longest_len)\n                for child in graph[cur_node]: \n                    to_visit.append((child,depth+1))\n            \n        return longest_len'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'longestStrChain'
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: 8279