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>
using namespace std;
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - 48;
    ch = getchar();
  }
  return x * f;
}
template <typename T>
inline bool Max(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool Min(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
long long n, m, k, x = -1, y, fg = 0;
int main() {
  n = gi(), m = gi(), k = gi();
  if (2 * n * m % k != 0) {
    return puts("NO"), 0;
  }
  if (n > m) swap(n, m), fg = 1;
  long long s = 2 * n * m / k, st = max(1ll, s / m), up = min(n, s);
  for (long long i = st; i <= up; ++i)
    if (s % i == 0 && s / i <= m) {
      x = i, y = s / i;
      goto A;
    }
A:;
  if (x == -1) return puts("NO"), 0;
  puts("YES");
  puts("0 0");
  if (!fg) {
    printf("%I64d 0\n", x);
    printf("%I64d %I64d\n", x, y);
  } else {
    printf("0 %I64d\n", x);
    printf("%I64d %I64d\n", y, x);
  }
  return 0;
}
