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.Scanner;

public class ProblemD2 {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		long n = scan.nextInt();
		long m = scan.nextInt();
		long k = scan.nextInt();
		scan.close();
		if(2 * n * m % k != 0) {
			System.out.println("NO");
			System.exit(0);
		}
		if(k % 2 == 0) {
			k = k / 2;
			long g = gcd(n, k);
			long x = n / g;
			long t = k / g;
			long y = m / t;
			if(x <= n && y <= m) {
				System.out.println("YES");
				System.out.println(0 + " " + 0);
				System.out.println(x + " " + 0);
				System.out.println(0 + " " + y);
			}else if(x <= n && y <= m){
				System.out.println("YES");
				System.out.println(0 + " " + 0);
				System.out.println(x + " " + 0);
				System.out.println(0 + " " + y);
			}else {
				System.out.println("NO");
			}
		}else {
			long g = gcd(n, k);
			long x = n / g;
			long t = k / g;
			long y = m / t;
			if(2 * x <= n && y <= m) {
				System.out.println("YES");
				System.out.println(0 + " " + 0);
				System.out.println(2 * x + " " + 0);
				System.out.println(0 + " " + y);
			}else if(x <= n && 2 * y <= m){
				System.out.println("YES");
				System.out.println(0 + " " + 0);
				System.out.println(x + " " + 0);
				System.out.println(0 + " " + 2 * y);
			}else {
				System.out.println("NO");
			}
		}

	}
	static long gcd(long m, long n) {
	    if(m < n) return gcd(n, m);
	    if(n == 0) return m;
	    return gcd(n, m % n);
	}
}