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;
long long n, m, k;
long long x;
void factor(long long s) {
  long long st = 1;
  if (s > 1000000000000000) st = s / max(n, m);
  for (long long i = st; i * i <= s; i++) {
    if (s % i == 0) {
      if ((i <= n && x / i <= m) || (i <= m && x / i <= n)) {
        cout << "YES" << '\n';
        if (i <= n && x / i <= m) {
          cout << 0 << ' ' << 0 << '\n';
          cout << 0 << ' ' << x / i << '\n';
          cout << i << ' ' << 0 << '\n';
        } else {
          cout << 0 << ' ' << 0 << '\n';
          cout << 0 << ' ' << i << '\n';
          cout << x / i << ' ' << 0 << '\n';
        }
        exit(0);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  long long c = k;
  if (k % 2 == 0) c /= 2;
  if ((n * m) % c)
    cout << "NO";
  else {
    x = (n * m * 2) / k;
    factor(x);
    cout << "NO";
  }
  return 0;
}
