In a row of trees, the i-th tree produces fruit with type tree[i].
You start at any tree of your choice, then repeatedly perform the following steps:

Add one piece of fruit from this tree to your baskets.  If you cannot, stop.
Move to the next tree to the right of the current tree.  If there is no tree to the right, stop.

Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.
You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.
What is the total amount of fruit you can collect with this procedure?
 
Example 1:
Input: [1,2,1]
Output: 3
Explanation: We can collect [1,2,1].


Example 2:
Input: [0,1,2,2]
Output: 3
Explanation: We can collect [1,2,2].
If we started at the first tree, we would only collect [0, 1].


Example 3:
Input: [1,2,3,2,2]
Output: 4
Explanation: We can collect [2,3,2,2].
If we started at the first tree, we would only collect [1, 2].


Example 4:
Input: [3,3,3,1,2,1,1,2,3,3,4]
Output: 5
Explanation: We can collect [1,2,1,1,2].
If we started at the first tree or the eighth tree, we would only collect 4 fruits.

 



Note:

1 <= tree.length <= 40000
0 <= tree[i] < tree.length

```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 = '# 问题可以抽象为给定一个数组，求解最多选择两种数字的情况下，最大的连续子序列长度，其中数组和原题目一样，每一个数字代表一个水果。\n\n# 我们可以使用滑动窗口来解决。 思路和 【1004. 最大连续 1 的个数 III】滑动窗口（Python3） 一样。\n\n\n\nclass Solution:\n    def atMostK(self, nums, K):\n        counter = collections.Counter()\n        res = i = 0\n        for j in range(len(nums)):\n            if counter[nums[j]] == 0:\n                K -= 1\n            counter[nums[j]] += 1  #然后加上1\n            while K < 0:\n                counter[nums[i]] -= 1  #移去最外面的，然后自己往左走\n                if counter[nums[i]] == 0:\n                    K += 1  #说明有一个品种多出来了，所以在没有到可以移出品种之前，需要一直变动\n                i += 1\n            res = max(res, j - i + 1)  #因为还要多一个\n        return res\n\n    def totalFruit(self, tree: List[int]) -> int:\n        return self.atMostK(tree, 2)   #最多两个\n\n# 复杂度分析\n\n# 时间复杂度：O(N)O(N)\n# 空间复杂度：O(N))O(N)) （我们使用了Counter）\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'totalFruit'
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: 6496