You are given an array $a$ of length $n$. A subsequence of this array is valid, if it satisfies these two conditions:
- There shouldn't be any two even numbers within a distance of $K$, both which have been chosen in the subsequence. i.e. there shouldn't be two indices $i, j$ such that $a_i$ and $a_j$ are even, $|i - j| \leq K$ and $a_i$ and $a_j$ are in the subsequence. 
- Similarly, there shouldn't be any two odd numbers within a distance of $K$, both which have been chosen in the subsequence
The sum of a subsequence is the sum of all the numbers in it. Your task is find the maximum sum possible in a valid subsequence of the given array. Print this maximum sum.

-----Input-----
- The first line of the input contains an integer $T$ denoting the number of test cases. The description of the test cases follows.
- The first line of each test case contains two space-separated integers $n, k$.
- The second line of each test case contains $n$ space-separated integers denoting the array $a$.

-----Output-----
For each test case, output an integer corresponding to the answer of the problem.

-----Constraints-----
- $1 \le T \le 10^5$
- $1 \le n \leq 10^5$
- $1 \le k \leq  n$
- $1 \le a_i \leq 10^9$
- Sum of $n$ over all the test cases doesn't exceed $10^6$

-----Example Input-----
3
1 1
3
2 1
2 2
5 2
1 2 3 4 6

-----Example Output-----
3
2
11

-----Explanation:-----
Testcase 2: Only one of the two 2s can be chosen. Hence the answer is 2.
Testcase 3: The subsequence containing the second, third and fifth numbers is a valid subsequence, and its sum is 2+3+6 = 11. You can check that this is the maximum possible, and hence is the answer.

```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

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "'''\n    ssisodia507@gmail.com\n'''\n\n\ndef solve(array,k,n):\n \n # storing even and odd values\n even=[]\n odd=[]\n for i in range(n):\n  if array[i]%2==0:\n   even.append(array[i])\n   odd.append(0)\n  else:\n   odd.append(array[i])\n   even.append(0)\n \n # dpEven will store the maximum sum of even elements till index i-th\n # and dpOdd of odd elements till index i-th\n dpEven=[0]*n\n dpOdd=[0]*n\n dpEven[0]=even[0]\n dpOdd[0]=odd[0]\n \n # from index 0 to k we can only select one even value and one odd value\n # so store the maximum even value till index k and maximum odd value \n # till index k\n i=1\n while i<=k and i<n:\n  dpEven[i]=max(dpEven[i-1],even[i])\n  dpOdd[i]=max(dpOdd[i-1],odd[i])\n  i+=1\n \n # now for any element we have two choices\n # either choose it or leave it.\n \n # we will consider the case which will give maximum sum\n # so we will find the sum after leaving the element(which is previous\n # maximum sum ) and after choosing the element(previous sum+currentvalue) \n # and pick the one which give maximum sum\n \n # we will handle even and odd cases separately \n for i in range(k+1,n):\n  dpEven[i]=max(dpEven[i-1],dpEven[i-k-1]+even[i])\n for i in range(k+1,n):\n  dpOdd[i]=max(dpOdd[i-1],dpOdd[i-k-1]+odd[i])\n return dpOdd[-1]+dpEven[-1]\nqueries=[]\nfor _ in range(int(input())):\n n,k=map(int,input().split( ))\n queries.append((list(map(int,input().split( ))),k))\n\nfor query in queries:\n print(solve(query[0],query[1],len(query[0])))"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13373