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;
long long n, m, k;
int main() {
  scanf("%lld %lld %lld", &n, &m, &k);
  if ((2 * n * m) % k) return printf("NO\n"), 0;
  long long p = 2 * n * m / k;
  long long L = 1, R = n, x = n;
  while (L <= R) {
    long long mid = (L + R) / 2;
    if (mid * m >= p)
      R = mid - 1, x = mid;
    else
      L = mid + 1;
  }
  if (p % x) {
    long long r;
    while (x <= n) {
      r = p / (p / x);
      if (p % r == 0) {
        x = r;
        break;
      }
      x = r + 1;
    }
  }
  if (x > n) return printf("NO\n"), 0;
  long long y = p / x;
  if (y > m) return printf("NO\n"), 0;
  printf("YES\n");
  printf("0 0\n");
  printf("%lld 0\n", x);
  printf("%lld %lld\n", x, y);
}
