There are N workers.  The i-th worker has a quality[i] and a minimum wage expectation wage[i].
Now we want to hire exactly K workers to form a paid group.  When hiring a group of K workers, we must pay them according to the following rules:

Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
Every worker in the paid group must be paid at least their minimum wage expectation.

Return the least amount of money needed to form a paid group satisfying the above conditions.
 



Example 1:
Input: quality = [10,20,5], wage = [70,50,30], K = 2
Output: 105.00000
Explanation: We pay 70 to 0-th worker and 35 to 2-th worker.


Example 2:
Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], K = 3
Output: 30.66667
Explanation: We pay 4 to 0-th worker, 13.33333 to 2-th and 3-th workers seperately. 

 
Note:

1 <= K <= N <= 10000, where N = quality.length = wage.length
1 <= quality[i] <= 10000
1 <= wage[i] <= 10000
Answers within 10^-5 of the correct answer will be considered correct.

```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 mincostToHireWorkers(self, quality: List[int], wage: List[int], K: int) -> float:\n        '''\n1- expect[i] = wage[i]/quality[i]: meaning i-th worker claims expect[i] money per each unit of its quality. Therefore, if expect[i] > expect[j], that means if we pay j-th worker quality[j]*expect[i] he/she would be more than happy and it's more than its minimal requested wage.\n\n2- Therefore, for k workers sorted by their expect values, if we pay each worker q[i]*expect[k], both rules are satisfied. The total needed money = (sum(q_1 + q_2 + ... + q_k) * expect[k]). Note that this is the minimum money for this k workers, since you have to pay the k-th worker at least q[k]*expect[k]. This part is very tricky, think of this in contrary, if you don't pay each of K workers with expected[k], let's say if you pick expected[k-1], you will not satisfy kth worker. So, the minimum expected pay per quality is expected[k] to satisfy all k workers. Every worker is paid in terms of quality ratio so we come up with a common pay per quality. \n\n3- To recap, we sort workers based on their expect values. Say we are at worker i and want to form a k-group and we already know it would cost sum*expect[i]. To pay the minimum money we should minimize the sum, which can be found using a maxHeap (to replace the max value with a smaller one) to keep the smaller q's as we move forward.\n        \n        '''\n        workers = sorted([float(w) / q, q] for w, q in zip(wage, quality))\n        maxHeap = []\n        res = float('inf')\n        qsum = 0\n        for expect, q in workers:\n            qsum += q\n            heapq.heappush(maxHeap, -q)\n            if len(maxHeap) > K:\n                qsum += heapq.heappop(maxHeap)\n            if len(maxHeap) == K:\n                res = min(res, expect*qsum)\n        return res\n            \n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'mincostToHireWorkers'
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: 6480