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 x, long long y) {
  while (x && y) {
    if (x > y)
      x %= y;
    else
      y %= x;
  }
  return x + y;
}
signed main() {
  ios_base::sync_with_stdio(0);
  long long n, m, k;
  cin >> n >> m >> k;
  bool hs2 = false;
  if (k == 1) {
    cout << "NO";
    return 0;
  }
  if (k % 2 == 0) {
    hs2 = true;
    k /= 2;
  }
  long long dn = gcd(n, k);
  long long dm = gcd(m, k / dn);
  if (dn * dm != k) {
    cout << "NO";
    return 0;
  }
  long long tn = n / dn, tm = m / dm;
  cout << "YES\n";
  if (hs2) {
    cout << "0 0\n";
    cout << tn << " 0\n";
    cout << "0 " << tm << "\n";
  } else {
    if (tn < n)
      tn *= 2;
    else
      tm *= 2;
    cout << "0 0\n";
    cout << tn << " 0\n";
    cout << "0 " << tm << "\n";
  }
  return 0;
}
