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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
long long nwd(long long a, long long b) { return b ? nwd(b, a % b) : a; }
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  long long d = nwd(n * (long long)m, k);
  int tmp = d;
  long long dn = nwd(tmp, n);
  int nd = n / dn;
  tmp /= dn;
  long long dm = nwd(tmp, m);
  int md = m / dm;
  k /= d;
  if (k > 2) {
    printf("NO\n");
    return 0;
  }
  if (k == 1) {
    if (2 * nd <= n)
      printf("YES\n0 0\n%d 0\n0 %d\n", 2 * nd, md);
    else if (2 * md <= m)
      printf("YES\n0 0\n%d 0\n0 %d\n", nd, 2 * md);
    else
      printf("NO\n");
  } else
    printf("YES\n0 0\n%d 0\n0 %d\n", nd, md);
  return 0;
}
