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 judge(long long x) {
  if (n % x == 0 && m % (k / x) == 0) return 1;
  return 0;
}
int main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  if (2 * n * m % k) {
    puts("NO");
    return 0;
  }
  if (k % 2 == 1) {
    for (long long i = 1; i * i <= k; i++)
      if (k % i == 0) {
        if (judge(i)) {
          puts("YES");
          puts("0 0");
          if (i >= 2) {
            printf("%lld 0\n", n / i * 2);
            printf("0 %lld\n", m / (k / i));
          } else {
            printf("%lld 0\n", n / i);
            printf("0 %lld\n", m / (k / i) * 2);
          }
          return 0;
        }
        if (judge(k / i)) {
          puts("YES");
          puts("0 0");
          if (k / i >= 2) {
            printf("%lld 0\n", n / (k / i) * 2);
            printf("0 %lld\n", m / i);
          } else {
            printf("%lld 0\n", n / (k / i));
            printf("0 %lld\n", m / i * 2);
          }
          return 0;
        }
      }
  } else {
    k /= 2;
    for (long long i = 1; i * i <= k; i++)
      if (k % i == 0) {
        if (judge(i)) {
          puts("YES");
          puts("0 0");
          printf("%lld 0\n", n / i);
          printf("0 %lld\n", m / (k / i));
          return 0;
        }
        if (judge(k / i)) {
          puts("YES");
          puts("0 0");
          printf("%lld 0\n", n / (k / i));
          printf("0 %lld\n", m / i);
          return 0;
        }
      }
  }
}
