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;
template <typename _T>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void put(long long x, long long y) { cout << x << " " << y << "\n"; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  long long N = n, M = m;
  long long g1 = gcd(n, k);
  n /= g1, k /= g1;
  long long g2 = gcd(m, k);
  m /= g2, k /= g2;
  if (k > 2) {
    cout << "NO\n";
    return 0;
  }
  if (k == 2) {
    cout << "YES\n";
    put(0, 0);
    put(n, 0);
    put(0, m);
    return 0;
  }
  if (k == 1) {
    if (2 * n <= N) {
      cout << "YES\n";
      put(0, 0);
      put(2 * n, 0);
      put(0, m);
    } else if (2 * m <= M) {
      cout << "YES\n";
      put(0, 0);
      put(n, 0);
      put(0, 2 * m);
    } else {
      cout << "NO\n";
      return 0;
    }
  }
  return 0;
}
