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 <class T>
ostream& operator<<(ostream& stream, const vector<T> v) {
  stream << "[ ";
  for (int i = 0; i < (int)v.size(); i++) stream << v[i] << " ";
  stream << "]";
  return stream;
}
long long fpow(long long x, long long p, long long m) {
  long long r = 1;
  for (; p; p >>= 1) {
    if (p & 1) r = r * x % m;
    x = x * x % m;
  }
  return r;
}
long long inv(long long a, long long b) {
  return a < 2 ? a : ((a - inv(b % a, a)) * b + 1) / a % b;
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  if (2 * n * m % k != 0) {
    cout << "NO\n";
    return 0;
  }
  long long gc = gcd(2 * n, k);
  long long a = 2 * n / gc;
  long long b = m / (k / gc);
  if (a > n) a /= 2, b *= 2;
  cout << "YES\n";
  cout << 0 << " " << 0 << "\n";
  cout << a << " " << 0 << "\n";
  cout << 0 << " " << b << "\n";
  return 0;
}
