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;
using lint = long long int;
template <class T = int>
using V = vector<T>;
template <class T = int>
using VV = V<V<T> >;
template <class T>
void assign(V<T>& v, int n, const T& a = T()) {
  v.assign(n, a);
}
template <class T, class... U>
void assign(V<T>& v, int n, const U&... u) {
  v.resize(n);
  for (auto&& i : v) assign(i, u...);
}
template <class T>
set<T> divisor(T n) {
  set<T> res;
  for (T i = 1; i * i <= n; i++)
    if (n % i == 0) res.insert(i), res.insert(n / i);
  return res;
}
int main() {
  cin.tie(NULL);
  ios::sync_with_stdio(false);
  lint n, m, k;
  cin >> n >> m >> k;
  if (k == 1) return cout << "NO" << '\n', 0;
  if (2 * n * m % k) return cout << "NO" << '\n', 0;
  cout << "YES" << '\n';
  lint s = 2 * n * m / k;
  auto x = divisor(k);
  for (auto&& i : x) {
    lint j = k / i;
    lint a, b;
    bool ok = false;
    if (!(2 * n % i) and !(m % j)) {
      a = 2 * n / i, b = m / j;
      if (a > n or b > m)
        ok = false;
      else
        ok = true;
    }
    if (!ok and !(n % i) and !(2 * m % j)) {
      a = n / i, b = 2 * m / j;
      if (a > n or b > m)
        ok = false;
      else
        ok = true;
    }
    if (!ok) continue;
    cout << 0 << ' ' << 0 << '\n';
    cout << a << ' ' << 0 << '\n';
    cout << 0 << ' ' << b << '\n';
    return 0;
  }
}
