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.
#include <bits/stdc++.h>
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  if (2ll * n * m % k != 0)
    puts("NO");
  else {
    if (k % 2 == 0) {
      int g = gcd(n, k / 2);
      printf("YES\n0 0\n%d 0\n0 %d\n", n / g, m / (k / 2 / g));
    } else {
      bool flag = 1;
      int g = gcd(n, k);
      int x = n / g, y = m / (k / g);
      if (2 * x <= n)
        x *= 2;
      else if (2 * y <= m)
        y *= 2;
      else
        puts("NO"), flag = 0;
      if (flag) printf("YES\n0 0\n%d 0\n0 %d\n", x, y);
    }
  }
  return 0;
}
