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;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  long long g = gcd(n, k);
  long long n1 = n, m1 = m, k1 = k;
  n1 /= g;
  k1 /= g;
  g = gcd(m, k1);
  m1 /= g;
  k1 /= g;
  if (k1 == 1ll) {
    if (n1 * 2ll <= n)
      n1 *= 2ll;
    else if (m1 * 2ll <= m)
      m1 *= 2ll;
    else {
      printf("NO");
      return 0;
    }
  } else if (k1 != 2ll) {
    printf("NO");
    return 0;
  }
  printf("YES\n%lld 0\n0 0\n0 %lld", n1, m1);
}
