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>
inline int read() {
  char ch = getchar();
  int ret = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  for (; ch >= '0' && ch <= '9'; ch = getchar()) ret = ret * 10 + ch - '0';
  return ret * f;
}
using namespace std;
const int INF = 1 << 30;
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  if (n * m * 2 % k) {
    puts("NO");
    return 0;
  }
  k = n * m * 2 / k;
  long long t = sqrt(k);
  for (long long i = max(k / max(n, m), 1ll); i <= t; ++i)
    if (k % i == 0) {
      long long tmp = k / i;
      if (i <= n && tmp <= m) {
        printf("YES\n%d %d\n%I64d %d\n%d %I64d\n", 0, 0, i, 0, 0, tmp);
        return 0;
      }
      if (i <= m && tmp <= n) {
        printf("YES\n%d %d\n%I64d %d\n%d %I64d\n", 0, 0, tmp, 0, 0, i);
        return 0;
      }
    }
  puts("NO");
  return 0;
}
