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>
#pragma GCC optimize("Ofast")
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  if ((2ll * n * m) % k) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  cout << 0 << " " << 0 << '\n';
  long long a = (2ll * n * m) / k;
  vector<long long> v;
  n *= 2;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i) continue;
    v.push_back(i);
    if (i * i < n) v.push_back(n / i);
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  for (long long i : v) {
    if (i * 2 > n) continue;
    if (a % i == 0) {
      cout << 0 << " " << a / i << '\n';
      cout << i << " " << 0 << '\n';
      return 0;
    }
  }
  return 0;
}
