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;
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  if ((2LL * n * 1LL * m) % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  int x, y;
  bool b = false;
  if ((2 * n) % k == 0) {
    x = (2 * n) / k;
    y = m;
    b = true;
  }
  if ((2 * m) % k == 0) {
    x = n;
    y = (2 * m) / k;
    b = true;
  }
  for (int i = 2; i * i <= k && !b; i++) {
    if (k % i == 0) {
      int k1 = i;
      int k2 = k / i;
      if ((2 * n) % k1 == 0 && m % k2 == 0) {
        x = (2 * n) / k1;
        y = m / k2;
        b = true;
      } else if ((2 * n) % k2 == 0 && m % k1 == 0) {
        x = (2 * n) / k2;
        y = m / k1;
        b = true;
      }
    }
  }
  if (b) {
    cout << "YES" << endl;
    cout << "0 0" << endl;
    cout << x << " 0" << endl;
    cout << "0 " << y << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}
