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, nn, mm;
void pan(long long x, long long y) {
  printf("%s\n", "YES");
  printf("%d %d\n", 0, 0);
  printf("%lld %d\n", x, 0);
  printf("%lld %lld", x, y);
  exit(0);
}
long long gcd(long long x, long long y) {
  long long z = x % y;
  while (z) {
    x = y;
    y = z;
    z = x % y;
  }
  return y;
}
void find() {
  long long t = gcd(n, k);
  n /= t;
  k /= t;
  t = gcd(m, k);
  m /= t, k /= t;
  if (k == 1) {
    if (n * 2 <= nn)
      n *= 2;
    else
      m *= 2;
  }
  pan(n, m);
}
int main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  nn = n, mm = m;
  if (n * m * 2 % k != 0 || k == 1) {
    printf("%s\n", "NO");
  } else {
    find();
  }
}
