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, area;
long long f[100100];
int main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  if (2 * n * m % k != 0) {
    printf("NO\n");
    return 0;
  }
  area = 2 * n * m / k;
  long long p = area;
  for (long long i = 2; i * i <= p; i++)
    while (p % i == 0) f[++f[0]] = i, p /= i;
  if (p) f[++f[0]] = p;
  long long Max = max(n, m);
  p = 1;
  for (int i = 1; i <= f[0]; i++) {
    p *= f[i];
    if (p > Max) {
      p /= f[i];
      break;
    }
  }
  if (p <= n && area / p <= m)
    printf("YES\n0 0\n%lld 0\n0 %lld", p, area / p);
  else if (p <= m && area / p <= n)
    printf("YES\n0 0\n0 %lld\n%lld 0", p, area / p);
  else
    printf("YES\n0 0\n0 %lld\n%lld 0", m, area / m);
  return 0;
}
