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 maxn = 2e5 + 5;
long long cal(long long n, long long m, long long d) {
  long long a = (d + m - 1) / m, b = n;
  if (!a) a++;
  long long x;
  if (a > b) return -1;
  if (d % a == 0)
    return a;
  else if (d % b == 0)
    return b;
  while (a <= b) {
    a = d / (d / a);
    if (a > b) return -1;
    if (d % a == 0) return a;
    a++;
  }
  return -1;
}
int main() {
  long long n, m, k;
  scanf("%lld%lld%lld", &n, &m, &k);
  long long x, y;
  if (2 * n * m % k != 0 || (x = cal(n, m, 2 * n * m / k)) == -1)
    return printf("NO\n"), 0;
  y = 2 * n * m / k / x;
  printf("YES\n0 0\n0 %lld\n%lld 0\n", y, x);
}
