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;
const int maxn = 1e5 + 5;
const int maxm = 1e5 + 5;
const int maxl = 26;
long long n, m, k;
struct Edge {
  long long u, v, w;
  Edge(long long _u, long long _v, long long _w) : u(_u), v(_v), w(_w) {}
};
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n >> m >> k) {
    int flag = 0;
    if (k % 2) {
      if (n * m % k != 0) {
        flag = 1;
        cout << "NO" << endl;
      }
    } else if (n * m % k != 0 && n * m % (k / 2)) {
      flag = 1;
      cout << "NO" << endl;
    }
    if (flag) continue;
    cout << "YES" << endl;
    k = 2 * n * m / k;
    long long ans = k / n;
    if (k % n) {
      ans++;
    }
    long long ans2 = ans * n - k;
    cout << "0 0" << endl << n << " 1" << endl << ans2 << " " << ans << endl;
  }
}
