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>
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0)
    return (a + b);
  else
    return gcd((a < b ? a : b), (a > b ? a : b) % (a < b ? a : b));
}
int main() {
  long long x_limit, y_limit, k;
  scanf("%I64d %I64d %I64d", &x_limit, &y_limit, &k);
  long long g = gcd(x_limit, k);
  long long x = x_limit / g;
  k /= g;
  g = gcd(y_limit, k);
  long long y = y_limit / g;
  k /= g;
  if (k == 1) {
    if (2 * x <= x_limit)
      x = 2 * x;
    else
      y = 2 * y;
  }
  if (k <= 2) {
    printf("YES\n");
    printf("0 0\n%I64d 0\n0 %I64d\n", x, y);
  } else {
    printf("NO\n");
  }
  return 0;
}
