description: The only difference between this problem and D2 is that you don't have to provide the way to construct the answer in this problem, but you have to do it in D2.
  
  There's a table of n × m cells (n rows and m columns). The value of n ⋅ m is even.
  
  A domino is a figure that consists of two cells having a common side. It may be horizontal (one of the cells is to the right of the other) or vertical (one of the cells is above the other).
  
  You need to find out whether it is possible to place nm/2 dominoes on the table so that exactly k of them are horizontal and all the other dominoes are vertical. The dominoes cannot overlap and must fill the whole table.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10) — the number of test cases. Then t test cases follow.
  
  Each test case consists of a single line. The line contains three integers n, m, k (1 ≤ n,m ≤ 100, 0 ≤ k ≤ nm/2, n ⋅ m is even) — the number of rows, columns and horizontal dominoes, respectively.
  
  Output
  
  For each test case output "YES", if it is possible to place dominoes in the desired way, or "NO" otherwise.
  
  You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
  
  Example
  
  Input
  
  
  8
  4 4 2
  2 3 0
  3 2 3
  1 2 0
  2 4 2
  5 2 2
  2 17 16
  2 1 1
  
  
  Output
  
  
  YES
  YES
  YES
  NO
  YES
  NO
  YES
  NO
solution: from itertools import product, permutations, combinations, accumulate, compress
  #from collections import Counter, defaultdict, deque # deque([iterable[, maxlen]]) #appendleft popleft rotate
  #from heapq import heapify, heappop, heappush # func(heapifiedlist, item)
  #from bisect import bisect_left, bisect_right, insort # func(sortedlist, item)
  # TODO: more coping mechanisms...?
  #from sys import setrecursionlimit
  
  import os, sys
  input = lambda: sys.stdin.buffer.readline().rstrip(b'\r\n')
  
  def dbp(*args, **kwargs): # calling with dbp(locals()) is perfectly cromulent
  	print(*args, file=sys.stderr, **kwargs)
  
  a_str = lambda: input().decode('ascii')
  get_int_list = lambda: [int(x) for x in input().split()]
  join_str = lambda l, ch=' ': ch.join((str(x) for x in l))  # ch is a default, not kwarg
  
  def wa4(h, w, horzs, verts):
  	if h % 2 == 1: # odd height, even width, "h*w is even"
  		horzs -= w//2
  		if horzs < 0 or horzs % 2 == 1 or verts % 2 == 1:
  			return 'NO'
  		else:
  			return 'YES'
  
  	elif w % 2 == 1: # even height, odd width
  		verts -= h // 2
  		if verts < 0 or horzs % 2 == 1 or verts % 2 == 1:
  			return 'NO'
  		else:
  			return 'YES'
  
  	else: # both even
  		return 'YES' if horzs % 2 == 0 == verts % 2 else 'NO'
  
  
  
  def do_thing():
  	h, w, k = get_int_list()
  	verts = (h*w//2)-k
  	dbp('case:', h, w, k, verts)
  	return wa4(h, w, k, verts)
  
  def multicase():
  	maxcc = int(input().strip())
  	for cc in range(maxcc):
  		print(do_thing())
  
  if __name__ == "__main__":
  	multicase()
  	#print(do_thing())