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;
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  if (2 * n * m % k != 0) {
    cout << "NO" << '\n';
    return;
  }
  long long req = 2 * n * m / k;
  if (k < 2) {
    cout << "NO\n";
    return;
  }
  long long x1 = 0, y1 = 0;
  long long y2 = 0, x3 = n;
  for (long long i = max(1ll, req / max(n, m)); i * i <= req && i <= max(n, m);
       i++) {
    if (req % i == 0) {
      long long a = i;
      long long b = req / i;
      if (a <= n && b <= m) {
        cout << "YES\n";
        cout << x1 << ' ' << y1 << '\n';
        cout << a << ' ' << y2 << '\n';
        cout << x3 << ' ' << b << '\n';
        return;
      }
      if (a <= m && b <= n) {
        cout << "YES\n";
        cout << x1 << ' ' << y1 << '\n';
        cout << b << ' ' << y2 << '\n';
        cout << x3 << ' ' << a << '\n';
        return;
      }
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
