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 long long read(void) {
  long long x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9';
       x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar())
    ;
  return x * f;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  long long n = read(), m = read(), k = read();
  if (n * m * 2 % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  cout << "0 0" << endl;
  long long x = gcd(2 * n, k), y = k / x;
  x = 2 * n / x, y = m / y;
  if (x <= n && y <= m) {
    cout << x << " 0" << endl;
    cout << "0 " << y << endl;
  } else {
    x >>= 1, y <<= 1;
    cout << x << " 0" << endl;
    cout << "0 " << y << endl;
  }
  return 0;
}
