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 GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void finish(int a, int b) {
  printf("YES\n0 0 %d 0 0 %d", a, b);
  exit(0);
}
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  long long top = 2LL * n * m;
  long long bot = k;
  if (top % bot != 0) return !printf("NO\n");
  function<int(int, int)> gcd = [&](int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  };
  int g1 = gcd(2 * n, k);
  int a, b;
  a = 2 * n / g1;
  b = m / (k / g1);
  if (a <= n && b <= m) finish(a, b);
  g1 = gcd(2 * m, k);
  b = 2 * m / g1;
  a = n / (k / g1);
  if (a <= n && b <= m) finish(a, b);
  return 0;
}
