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>
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
int main() {
  long long int n, m, k, a, b, c;
  scanf("%lld%lld%lld", &n, &m, &k);
  a = n * m * 2;
  if (a % k != 0) {
    printf("NO\n");
    return 0;
  } else {
    if (k % 2 == 0) {
      k /= 2;
      a = gcd(m, k);
      k /= a;
      a = m / a;
      printf("YES\n");
      printf("0 0\n");
      printf("0 %lld\n", a);
      printf("%lld 0\n", n / k);
    } else {
      n *= 2;
      a = gcd(n, k);
      c = k / a;
      a = n / a;
      if (a <= n / 2) {
        printf("YES\n");
        printf("0 0\n");
        printf("0 %lld\n", m / c);
        printf("%lld 0\n", a);
      } else {
        n /= 2;
        m *= 2;
        a = gcd(m, k);
        k /= a;
        a = m / a;
        if (a <= m / 2) {
          printf("YES\n");
          printf("0 0\n");
          printf("0 %lld\n", a);
          printf("%lld 0\n", n / k);
        } else
          printf("NO\n");
      }
    }
  }
}
