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.
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskA {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    String[] s = br.readLine().split(" ");
    long n = Long.parseLong(s[0]);
    long m = Long.parseLong(s[1]);
    long k = Long.parseLong(s[2]);
    long prod = n*m;
    List<Long> factors = primeFactorization(k);
    long n2 = n;
    long m2 = m;
    boolean possible = true;
    boolean twoSkipped = false;
    for (long f:factors) {
      if (f==2 && !twoSkipped) {
        twoSkipped = true;
        continue;
      }
      if (n2%f==0) {
        n2 /= f;
      } else if (m2%f==0) {
        m2 /= f;
      } else {
        possible = false;
        break;
      }
    }
    if (possible) {
      if (!twoSkipped && n2*2<=n) {
        n2 *= 2;
      } else if (!twoSkipped && m2*2<=m) {
        m2 *= 2;
      }
      System.out.println("YES");
      System.out.println("0 0");
      System.out.println(n2+" 0");
      System.out.println("0 "+m2);
    } else {
      System.out.println("NO");
    }
  }
  static List<Long> primeFactorization(long n) {
    List<Long> factors = new ArrayList<>();
    if (n <= 1) return factors;
    Queue<Long> queue = new PriorityQueue<>(); queue.add(n);
    while (!queue.isEmpty()) {
      long divisor = queue.remove();
      if (isPrime(divisor)) factors.add(divisor);
      else {
        long next = pollardRho(divisor);
        if (next != divisor) queue.add(divisor / next);
        queue.add(next); } }
    return factors; }
  static long pollardRho(long n) {
    if (n % 2 == 0) return 2;
    long x = 2 + (long) (999999 * Math.random()); long y = x;
    long c = 2 + (long) (999999 * Math.random()); long d = 1;
    while (d == 1) {
      x = (x * x + c) % n; y = (y * y + c) % n; y = (y * y + c) % n;
      d = gcd(Math.abs(x - y), n);
      if (d == n) break; }
    return d; }
  static long gcd(long a, long b) {
    return b == 0 ? (a<0?-a:a) : gcd(b, a % b); }
  static boolean isPrime(long n) {
    if (n < 2) return false; if (n == 2 || n == 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    long limit = (long) Math.sqrt(n);
    for (long i = 5; i <= limit; i += 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true; }

}
