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) {
  while (y) x %= y, swap(x, y);
  return x;
}
pair<long long, long long> Try(long long a, long long b, long long ua,
                               long long ub, long long k) {
  long long a_gcd = Gcd(a, k);
  a /= a_gcd;
  k /= a_gcd;
  b /= k;
  if (a <= ua && b <= ub)
    return make_pair(a, b);
  else
    return make_pair(-1, -1);
}
int main() {
  cin >> n >> m >> k;
  if ((2 * n * m) % k != 0) {
    printf("NO\n");
  } else {
    auto ab1 = Try(2 * n, m, n, m, k);
    auto ba = Try(2 * m, n, m, n, k);
    auto ab2 = make_pair(ba.second, ba.first);
    if (ab1.first == -1) ab1 = ab2;
    if (ab1.first == -1)
      printf("NO\n");
    else {
      printf("YES\n");
      printf("0 0\n");
      printf("%d 0\n", ab1.first);
      printf("0 %d\n", ab1.second);
    }
  }
  return 0;
}
