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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
long long n, m, k;
int main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  if ((2 * n * m) % k != 0) {
    printf("NO\n");
    return 0;
  }
  long long xsize = n;
  long long ysize = m;
  int extra = 1;
  for (long long d = 2; d * d <= k; d++) {
    if (k % d == 0) {
      while (k % d == 0) {
        if (xsize % d != 0 && ysize % d != 0) {
          if (d == 2 && extra == 1) {
            extra = 0;
          } else {
            printf("NO\n");
            return 0;
          }
        } else if (xsize % d == 0) {
          xsize /= d;
        } else if (ysize % d == 0) {
          ysize /= d;
        }
        k /= d;
      }
    }
  }
  if (k > 1) {
    long long d = k;
    if (xsize % d != 0 && ysize % d != 0) {
      if (d == 2 && extra == 1) {
        extra = 0;
      } else {
        printf("NO\n");
        return 0;
      }
    } else if (xsize % d == 0) {
      xsize /= d;
    } else if (ysize % d == 0) {
      ysize /= d;
    }
    k /= d;
  }
  if (extra) {
    if (xsize * 2 <= n)
      xsize *= 2;
    else
      ysize *= 2;
  }
  printf("YES\n%lld %lld\n%lld %lld\n%lld %lld\n", 0LL, 0LL, xsize, 0LL, 0LL,
         ysize);
  return 0;
}
