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: //template
  
  import java.util.*;
  import java.io.*;
  public class pa{
      //SOLUTION BEGIN
      
      void pre() throws Exception{}
     
      void solve(int TC) throws Exception{
         int n=ni();
         int m=ni();
         int k=ni();
        int min=0,max=(n*m)/2;
        if(n%2==1) {
      	  min=m/2;
        }
        if(m%2==1) {
      	  max=n*(m-1)/2;
        }
        if(k>=min && k<=max && k%2==min%2) {
      	  pn("YES");
        }
        else {
      	  pn("NO");
        }
      }
      
      //SOLUTION END
      static long N = 1000000007L;
      void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
      static boolean multipleTC = true;
      FastReader in;PrintWriter out;
      void run() throws Exception{
          in = new FastReader();
          out = new PrintWriter(System.out);
          
          int T = (multipleTC)?ni():1;
          pre();for(int t = 1; t<= T; t++)solve(t);
          out.flush();
          out.close();
      }
      public static void main(String[] args) throws Exception{
          new pa().run();
      }
      int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
      void p(Object o){out.print(o);}
      void pn(Object o){out.println(o);}
      void pni(Object o){out.println(o);out.flush();}
      String n()throws Exception{return in.next();}
      String nln()throws Exception{return in.nextLine();}
      int ni()throws Exception{return Integer.parseInt(in.next());}
      long nl()throws Exception{return Long.parseLong(in.next());}
      double nd()throws Exception{return Double.parseDouble(in.next());}
  
      class FastReader{
          BufferedReader br;
          StringTokenizer st;
          public FastReader(){
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          public FastReader(String s) throws Exception{
              br = new BufferedReader(new FileReader(s));
          }
  
          String next() throws Exception{
              while (st == null || !st.hasMoreElements()){
                  try{
                      st = new StringTokenizer(br.readLine());
                  }catch (IOException  e){
                      throw new Exception(e.toString());
                  }
              }
              return st.nextToken();
          }
  
          String nextLine() throws Exception{
              String str = "";
              try{   
                  str = br.readLine();
              }catch (IOException e){
                  throw new Exception(e.toString());
              }  
              return str;
          }
      }
  	    
  	
  	
  	static long expo(long base, long exp)
  	{
  	    if (exp == 0)
  	        return 1;
  	
  	    if (exp == 1)
  	        return base % N;
  	
  	    long t = expo(base, exp / 2);
  	    t = (t * t) % N;
  	
  	    // if exponent is even value
  	    if (exp % 2 == 0)
  	        return t;
  	
  	    // if exponent is odd value
  	    else
  	        return ((base % N) * t) % N;
  	}
      
  }