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 = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
long long n, m, k, a;
long long gcd(long long a, long long b) {
  long long tmp;
  while (b) {
    tmp = b;
    b = a % b;
    a = tmp;
  }
  return a;
}
int main() {
  cin >> n >> m >> k;
  a = 2 * n * m / k;
  if (2 * n * m % k == 0) {
    puts("YES");
    long long a, b, g;
    if (k & 1) {
      g = gcd(n, k);
      a = n / g;
      k /= g;
      b = m / k;
      if (a * 2 < n)
        a *= 2;
      else
        b *= 2;
    } else {
      k /= 2;
      g = gcd(n, k);
      a = n / g;
      k /= g;
      b = m / k;
    }
    puts("0 0");
    printf("%I64d 0\n", a);
    printf("0 %I64d\n", b);
  } else {
    puts("NO");
  }
  return 0;
}
