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;
constexpr int MOD = 1000000007;
constexpr int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, -1, 1};
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "[";
  for (const auto &v : vec) {
    os << v << ",";
  }
  os << "]";
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
void solve() {
  long long N, M, K;
  cin >> N >> M >> K;
  for (int i = 1; i * i <= K; i++) {
    if (K % i == 0) {
      int j = K / i;
      if ((2 * N) % i == 0 && M % j == 0 && (2 * N) / i <= N) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << (2 * N) / i << " 0" << endl;
        cout << "0 " << M / j << endl;
        return;
      }
      if (N % i == 0 && (M * 2) % j == 0 && (2 * M) / j <= M) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << N / i << " 0" << endl;
        cout << "0 " << (2 * M) / j << endl;
        return;
      }
      if ((2 * N) % j == 0 && M % i == 0 && (2 * N) / j <= N) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << (2 * N) / j << " 0" << endl;
        cout << "0 " << M / i << endl;
        return;
      }
      if (N % j == 0 && (M * 2) % i == 0 && (2 * M) / i <= M) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << N / j << " 0" << endl;
        cout << "0 " << (2 * M) / i << endl;
        return;
      }
    }
  }
  cout << "NO" << endl;
}
int main() {
  std::cin.tie(0);
  std::ios::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(16);
  solve();
  return 0;
}
