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: #!/usr/bin/env python3
  import sys
  import getpass  # not available on codechef
  import math, random
  import functools, itertools, collections, heapq, bisect
  from collections import Counter, defaultdict, deque
  input = sys.stdin.readline  # to read input quickly
  
  # available on Google, AtCoder Python3, not available on Codeforces
  # import numpy as np
  # import scipy
  
  M9 = 10**9 + 7  # 998244353
  yes, no = "YES", "NO"
  # d4 = [(1,0),(0,1),(-1,0),(0,-1)]
  # d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
  # d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
  MAXINT = sys.maxsize
  
  # if testing locally, print to terminal with a different color
  OFFLINE_TEST = getpass.getuser() == "hkmac"
  # OFFLINE_TEST = False  # codechef does not allow getpass
  def log(*args):
      if OFFLINE_TEST:
          print('\033[36m', *args, '\033[0m', file=sys.stderr)
  
  def solve(*args):
      # screen input
      if OFFLINE_TEST:
          log("----- solving ------")
          log(*args)
          log("----- ------- ------")
      return solve_(*args)
  
  def read_matrix(rows):
      return [list(map(int,input().split())) for _ in range(rows)]
  
  def read_strings(rows):
      return [input().strip() for _ in range(rows)]
  
  def minus_one(arr):
      return [x-1 for x in arr]
  
  def minus_one_matrix(mrr):
      return [[x-1 for x in row] for row in mrr]
  
  # ---------------------------- template ends here ----------------------------
  
  
  def solve_(a,b,c):
      # your solution here
  
      required = 0
      blocked = 0
      if a%2:
          required = b//2
      
      if b%2:
          blocked = a//2
  
      log(required)
      remaining = c-required
  
      if remaining < 0:
          return no
      
      if remaining%2:
          return no
  
      if remaining > (a*b)//2 - blocked:
          return no
  
      return yes
  
  
  # for case_num in [0]:  # no loop over test case
  # for case_num in range(100):  # if the number of test cases is specified
  for case_num in range(int(input())):
  
      # read line as an integer
      # k = int(input())
  
      # read line as a string
      # srr = input().strip()
  
      # read one line and parse each word as a string
      # lst = input().split()
      
      # read one line and parse each word as an integer
      a,b,c = list(map(int,input().split()))
      # lst = list(map(int,input().split()))
      # lst = minus_one(lst)
  
      # read multiple rows
      # arr = read_strings(k)  # and return as a list of str
      # mrr = read_matrix(k)  # and return as a list of list of int
      # mrr = minus_one_matrix(mrr)
  
      res = solve(a,b,c)  # include input here
  
      # print length if applicable
      # print(len(res))
  
      # parse result
      # res = " ".join(str(x) for x in res)
      # res = "\n".join(str(x) for x in res)
      # res = "\n".join(" ".join(str(x) for x in row) for row in res)
  
      # print result
      # print("Case #{}: {}".format(case_num+1, res))   # Google and Facebook - case number required
  
      print(res)