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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m, k;
  cin >> n >> m >> k;
  long double r = (n * m) / ((long double)k);
  if (r - floor(r) != 0 && r - floor(r) != 0.5 ||
      r > n * m / ((long double)2)) {
    cout << "NO" << endl;
    return 0;
  }
  r *= 2;
  long long int xx = r;
  long long int yy = 1;
  long long int mlt = r / max(n, m) - 1;
  do {
    mlt++;
    if (xx % mlt != 0) continue;
    long long int x = xx / mlt;
    long long int y = yy * mlt;
    if (x <= n && y <= m) {
      cout << "YES" << endl;
      cout << "0 0" << endl;
      cout << "0 " << y << endl;
      cout << x << " 0" << endl;
      return 0;
    } else if (x <= m && y <= n) {
      cout << "YES" << endl;
      cout << "0 0" << endl;
      cout << "0 " << x << endl;
      cout << y << " 0" << endl;
      return 0;
    }
    if (y > n && y > m) break;
    if (x < y) break;
  } while (true);
  cout << "NO" << endl;
  return 0;
}
