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 getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
long long n, m, k, t;
void pr(long long x, long long y) {
  puts("YES");
  cout << "0 0\n";
  cout << x << " 0\n";
  cout << "0 " << y << endl;
}
int main() {
  cin >> n >> m >> k;
  if (n * m * 2 % k != 0) return puts("NO"), 0;
  t = k * 2;
  for (long long i = 1; i * i <= t; ++i)
    if (t % i == 0) {
      if (i < 2 || t / i < 2) continue;
      if (n * 2 % i == 0 && m * 2 % (t / i) == 0)
        return pr(n * 2 / i, m * 2 / (t / i)), 0;
      if (n * 2 % (t / i) == 0 && m * 2 % i == 0)
        return pr(n * 2 / (t / i), m * 2 / i), 0;
    }
  puts("NO");
  return 0;
}
