Vasya has got three integers n, m and k. He'd like to find three integer points (x_1, y_1), (x_2, y_2), (x_3, y_3), such that 0 ≤ x_1, x_2, x_3 ≤ n, 0 ≤ y_1, y_2, y_3 ≤ m and the area of the triangle formed by these points is equal to nm/k.

Help Vasya! Find such points (if it's possible). If there are multiple solutions, print any of them.

Input

The single line contains three integers n, m, k (1≤ n, m ≤ 10^9, 2 ≤ k ≤ 10^9).

Output

If there are no such points, print "NO".

Otherwise print "YES" in the first line. The next three lines should contain integers x_i, y_i — coordinates of the points, one point per line. If there are multiple solutions, print any of them.

You can print each letter in any case (upper or lower).

Examples

Input

4 3 3


Output

YES
1 0
2 3
4 1


Input

4 4 7


Output

NO

Note

In the first example area of the triangle should be equal to nm/k = 4. The triangle mentioned in the output is pictured below: 

<image>

In the second example there is no triangle with area nm/k = 16/7.
import java.util.*;

public class GFG {
	public static void main (String[] args) {
	Scanner sc=new Scanner(System.in);
	long n=sc.nextLong(),m=sc.nextLong(),k=sc.nextLong();
    if(((2*n)*m)%k!=0)
      System.out.println("NO");
      else
      {   
          if((2*n)%k==0)
         { System.out.println("YES");
          System.out.println(0+" "+0);
             System.out.println(0+" "+m);
             System.out.println((2*n/k)+" "+0);}
             else if((2*m)%k==0)
             { System.out.println("YES");
          System.out.println(0+" "+0);
              System.out.println(n+" "+0);
             System.out.println(0+" "+(2*m/k));}   
            else{
                if((2*m*n/k)<=m)
                 { System.out.println("YES");
          System.out.println(0+" "+0);
              System.out.println(1+" "+0);
             System.out.println(0+" "+(2*m*n/k));
                     }  
                 else if((2*m*n/k)<=n)
                  { System.out.println("YES");
          System.out.println(0+" "+0);
              System.out.println(0+" "+1);
             System.out.println((2*m*n/k)+" "+0);
                     }  
               else
               {
                    long num1=k/2,num2=n,num3=m,num4=k/2;

        while (num1!=num2) {
        	if(num1>num2)
                num1=num1-num2;
            else
                num2=num2-num1;
        }
 while (num3!=num4) {
        	if(num3>num4)
                num3=num3-num4;
            else
                num4=num4-num3;
        }
        
             System.out.println("YES");
          System.out.println(0+" "+0);
              System.out.println(0+" "+(m/num3));
             System.out.println((n/num2)+" "+0);
             
        
               }
            }
      
   
	}
	}
}