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 gcd(long long a, long long b) {
  if (!a || !b) return (!a ? b : a);
  if (a < b) swap(a, b);
  long long f = a & 1, s = b & 1;
  if (f && s) {
    return gcd((a - b) >> 1, b);
  } else if (!f && s) {
    return gcd(a >> 1, b);
  } else if (f && !s) {
    return gcd(a, b >> 1);
  } else {
    return (gcd(a >> 1, b >> 1) << 1);
  }
}
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  if ((2 * n * m) % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  long long a = gcd(n, k);
  cout << "YES" << endl;
  if (2 * a * m / k <= m) {
    cout << 0 << " " << 0 << endl;
    cout << n / a << " " << 0 << endl;
    cout << 0 << " " << 2 * a * m / k << endl;
    return 0;
  }
  if (2 * n / a <= n) {
    cout << 0 << " " << 0 << endl;
    cout << 2 * n / a << " " << 0 << endl;
    cout << 0 << " " << a * m / k << endl;
  }
}
