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;
void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
map<long long, long long> dict;
bool func(long long n, long long x) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      dict[i] += x;
      if (n / i != i) {
        dict[n / i] += x;
      }
      if (dict[i] < 0 || dict[n / i] < 0) {
        return false;
      }
    }
  }
  return true;
}
void out(long long x, long long y) {
  cout << x << " 0\n";
  cout << "0 " << y << '\n';
}
signed main() {
  fast_io();
  long long n, m, k;
  cin >> n >> m >> k;
  if (2 * n * m / k != 2 * n * m / (double)k) {
    cout << "NO";
    return 0;
  }
  func(2 * n, 1);
  func(2 * m, 1);
  func(k, -1);
  cout << "YES\n0 0\n";
  long long prod = 2 * n * m / k;
  long long mx = max(n, m), mn = min(n, m);
  for (auto p : dict) {
    long long x = p.first;
    if (p.second <= 0 || prod % x) continue;
    if (x <= mn) {
      long long y = prod / x;
      if (y <= mx) {
        if (x <= n && y <= m) {
          out(x, y);
          return 0;
        } else {
          out(y, x);
          return 0;
        }
      }
    } else if (x <= mx) {
      long long y = prod / x;
      if (y <= mn) {
        if (x <= n && y <= m) {
          out(x, y);
          return 0;
        } else {
          out(y, x);
          return 0;
        }
      }
    }
  }
  return 0;
}
