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 print(long long a, long long b) {
  cout << "YES" << endl;
  cout << "0 0" << endl;
  cout << "0 " << b << endl;
  cout << a << " 0" << endl;
}
void ziggy(long long n, long long m, long long k) {
  bool f = k & 1;
  if (f) {
    for (long long i = 1; i * i <= k; i++)
      if (k % i == 0) {
        long long e = k / i;
        if (n % i == 0 && m % e == 0) {
          long long v1 = n / i, v2 = m / e;
          if (2 * v1 <= n) {
            print(2 * v1, v2);
            return;
          }
          if (2 * v2 <= m) {
            print(v1, 2 * v2);
            return;
          }
        }
        if (n % e == 0 && m % i == 0) {
          long long v1 = n / e, v2 = m / i;
          if (2 * v1 <= n) {
            print(2 * v1, v2);
            return;
          }
          if (2 * v2 <= m) {
            print(v1, 2 * v2);
            return;
          }
        }
      }
  } else {
    k /= 2;
    for (long long i = 1; i * i <= k; i++)
      if (k % i == 0) {
        long long e = k / i;
        if (n % i == 0 && m % e == 0) {
          print(n / i, m / e);
          return;
        }
        if (n % e == 0 && m % i == 0) {
          print(n / e, m / i);
          return;
        }
      }
  }
  cout << "NO" << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long A, n, k, m;
  cin >> n >> m >> k;
  if ((2 * n * m) % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  ziggy(n, m, k);
  return 0;
}
