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(object):\n       \n    \n    def mincostToHireWorkers(self, quality, wage, K):\n        from fractions import Fraction\n        workers = sorted((Fraction(w, q), q, w)\n                         for q, w in zip(quality, wage))\n\n        ans = float('inf')\n        pool = []\n        sumq = 0\n        for ratio, q, w in workers:\n            heapq.heappush(pool, -q)\n            sumq += q\n            \n           \n            if len(pool) > K:\n                sumq += heapq.heappop(pool)\n                \n            if len(pool) == K:\n                ans = min(ans, ratio * sumq)\n           \n \n        return float(ans)\n\n\n# Idea is as follows: If workers are paid in ratio of their qualities, then w_j = (w_ref/q_ref) * q_j\n\n# After paying in the above ratio, to ensure we pay workers their min wage, then we need to select the higher wage/ratio as reference among two possible wage/quality ratios \n\n\n# So if the size of array is k, then we just select the highest ratio and use that as ref, then \n#  min cost =  (highest ratio ) * (sum of all Qualities)\n\n# or in other words, sort the wage to ratios in increasing order, then last one will be ref ration and this ref ratio times sumq = ans.\n\n# now consider k+1 elements present. then same as above logic, if we sort the wage/quality ratios in increasing order, then any of the ratios from kth to later can be a candidate ref ratio. If k elements, we already know the ans is kth ratio * sumq. Now consider k+1 th element. We can have this as ref ration instead. Then which one would we remove to get k elements? We want to remove the element with highest quality as that would reduce the cost by max. (remember total cost = ref ratio * ( sum of qualities)). So we use max heap to maintain qualities and remove the highest quality when we consider all indexes > k. and among all those possible indexes k to n, we find which gives min cost.\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: 12009