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 n, m, k, x, y;
int gcd(int a, int b) {
  if (a < b) std::swap(a, b);
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  x = gcd(n, k);
  k /= x;
  x = n / x;
  y = gcd(m, k);
  k /= y;
  y = m / y;
  if (k != 1 && k != 2) return 0 * puts("NO");
  if (k == 1) {
    if (x * 2 <= n)
      x *= 2;
    else if (y * 2 <= m)
      y *= 2;
    else
      return 0 * puts("NO");
    puts("YES");
    puts("0 0");
    printf("0 %d\n", y);
    printf("%d 0\n", x);
  }
  if (k == 2) {
    puts("YES");
    puts("0 0");
    printf("0 %d\n", y);
    printf("%d 0\n", x);
  }
}
