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(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  long long xx = (2 * n * m);
  if (xx % k || k == 1) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  xx /= k;
  for (long long i = 1; i * i <= n; i++) {
    long long el = i;
    if (n % el == 0 && xx % el == 0) {
      long long ht = xx / el;
      if (ht <= m) {
        cout << "0 0" << endl << el << " 0" << endl << n << ' ' << ht;
        return 0;
      }
    }
    if (n % i != 0) {
      continue;
    }
    el = n / i;
    if (n % el == 0 && xx % el == 0) {
      long long ht = xx / el;
      if (ht <= m) {
        cout << "0 0" << endl << el << " 0" << endl << n << ' ' << ht;
        return 0;
      }
    }
  }
  for (long long i = 1; i * i <= m; i++) {
    long long el = i;
    if (m % el == 0 && xx % el == 0) {
      long long ht = xx / el;
      if (ht <= n) {
        cout << "0 0" << endl << ht << " 0" << endl << n << ' ' << el;
        return 0;
      }
    }
    if (m % i != 0) {
      continue;
    }
    el = m / i;
    if (m % el == 0 && xx % el == 0) {
      long long ht = xx / el;
      if (ht <= n) {
        cout << "0 0" << endl << ht << " 0" << endl << n << ' ' << el;
        return 0;
      }
    }
  }
  return 0;
}
