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;
int main() {
  int t;
  long long n, m, k;
  scanf("%I64d%I64d%I64d", &n, &m, &k);
  if ((n * m * 2) % k == 0) {
    if ((2 * m) % k == 0) {
      printf("YES\n0 0\n");
      printf("%I64d 0\n", n);
      printf("0 %I64d\n", 2 * m / k);
    } else if ((2 * n) % k == 0) {
      printf("YES\n0 0\n");
      printf("0 %I64d\n", m);
      printf("%I64d 0\n", 2 * n / k);
    } else if (k % (2 * m) == 0) {
      printf("YES\n0 0\n");
      printf("%I64d 0\n", 2 * n * m / k);
      printf("0 1\n");
    } else if (k % (2 * n) == 0) {
      printf("YES\n0 0\n");
      printf("0 %I64d\n", 2 * n * m / k);
      printf("1 0\n");
    } else {
      long long bb = 2 * n * m / k;
      if (n > m) {
        for (long long i = 1; i <= bb; i++) {
          if (bb % i == 0 && i <= n && bb / i <= m) {
            printf("YES\n0 0\n");
            printf("%I64d 0\n", i);
            printf("0 %I64d\n", bb / i);
            break;
          }
        }
      } else {
        for (long long i = 1; i <= bb; i++) {
          if (bb % i == 0 && i <= m && bb / i <= n) {
            printf("YES\n0 0\n");
            printf("0 %I64d\n", i);
            printf("%I64d 0\n", bb / i);
            break;
          }
        }
      }
    }
  } else {
    printf("NO\n");
  }
  return 0;
}
