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;
int32_t main() {
  long long n, m, k;
  scanf("%lld%lld%lld", &n, &m, &k);
  if ((2 * n * m) % k != 0)
    printf("NO\n");
  else {
    long long temp = 2 * n * m / k;
    for (int i = max(temp / max(n, m), 1ll); i * i <= temp && i <= n && i <= m;
         i++) {
      if (temp % i == 0) {
        long long x = temp / i, y = i;
        if (x <= n && y <= m) {
          printf("YES\n");
          printf("0 0\n");
          printf("%lld 0\n", x);
          printf("0 %lld\n", y);
          return 0;
        }
        if (x <= m && y <= n) {
          printf("YES\n");
          printf("0 0\n");
          printf("%lld 0\n", y);
          printf("0 %lld\n", x);
          return 0;
        }
      }
    }
    printf("NO\n");
  }
}
