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>
static ostream& operator<<(ostream& stream, const vector<T>& t);
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
template <typename T>
static istream& read(T, T, istream& = cin);
template <typename T>
static void print(T x, string end = "\n") {
  cout << x << end;
}
void no() {
  print("NO");
  exit(0);
}
void _() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long g = gcd(n * m, k);
  long long area_n = n * m / g;
  long long area_d = k / g;
  long long gn = gcd(g, n);
  long long gm = g / gn;
  if (area_d > 2) no();
  if (area_d == 1) {
    area_n *= 2;
    area_d *= 2;
    if (gn >= 2)
      n *= 2;
    else
      m *= 2;
  }
  n /= gn;
  m /= gm;
  print("YES");
  cout << n << " 0\n";
  cout << "0 " << m << '\n';
  cout << "0 0\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(15);
  _();
}
