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 dx[4] = {1, 0, -1, 0};
long long dy[4] = {0, -1, 0, 1};
template <class T>
void inputVector(vector<T>& v, long long n) {
  v.resize(n);
  for (long long i = 0; i < v.size(); i++) cin >> v[i];
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
signed main() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long val = 2 * n * m;
  if (val % k) {
    cout << "NO" << endl;
    return 0;
  }
  long long A = 2 * n;
  long long B = m;
  long long tmp = k;
  if (tmp > 1) {
    long long hoge = gcd(tmp, A);
    tmp /= hoge;
    A /= hoge;
  }
  if (tmp > 1) {
    long long hoge = gcd(tmp, B);
    tmp /= hoge;
    B /= hoge;
  }
  if (A > n) {
    if (A % 2 != 0) {
      cout << "ERROR" << endl;
      return 0;
    }
    A /= 2;
    B *= 2;
  }
  cout << "YES" << endl;
  cout << A << " 0" << endl;
  cout << "0 " << B << endl;
  cout << "0 0" << endl;
  cerr << "S   = " << A * B / 2.0 << endl;
  cerr << "val = " << (double)n * m / k << endl;
  return 0;
}
