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;
const int inf = 1 << 30;
int n, m, k;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  scanf("%d%d%d", &n, &m, &k);
  int n1 = n, m1 = m;
  int d = gcd(n, k);
  k /= d;
  n1 /= d;
  d = gcd(m, k);
  k /= d;
  m1 /= d;
  if (k > 2)
    puts("NO");
  else {
    if (k == 1) {
      if (n1 < n)
        n1 *= 2;
      else if (m1 < m)
        m1 *= 2;
      else {
        puts("NO");
        return 0;
      }
    }
    puts("YES");
    printf("0 0\n%d 0\n0 %d\n", n1, m1);
  }
  return 0;
}
