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;
int main() {
  cin >> n >> m >> k;
  long long s = n * m * 2;
  if (s % k || k == 1) return printf("NO"), 0;
  long long A = s / k, Min = min(n, m);
  printf("YES\n0 0\n");
  long long AA = sqrt(A);
  if (n == Min)
    for (long long i = A / m + !!(A % m); i <= n; ++i) {
      if (A % i == 0) {
        long long x = i, y = A / i;
        if (y <= m) return printf("%d 0\n", x), printf("0 %d", y), 0;
      }
    }
  if (m == Min)
    for (long long i = A / n + !!(A % n); i <= m; ++i) {
      if (A % i == 0) {
        long long x = i, y = A / i;
        if (y <= n) return printf("0 %d\n", x), printf("%d 0", y), 0;
      }
    }
  return 0;
}
