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;
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int main() {
  while (~scanf("%lld%lld%lld", &n, &m, &k)) {
    if (n * m * 2 % k == 0)
      printf("YES\n");
    else {
      printf("NO\n");
      continue;
    }
    long long s = 2 * n * m / k;
    printf("0 0\n");
    if (2 * m * n == k)
      printf("0 1\n1 0\n");
    else if (2 * n % k == 0)
      printf("%lld %lld\n%lld %lld\n", 2 * n / k, 0ll, 0ll, m);
    else if (2 * m % k == 0)
      printf("%lld %lld\n%lld %lld\n", 0ll, 2 * m / k, n, 0ll);
    else {
      long long ans = n * m / k;
      if (ans > max(n, m)) {
        long long x = gcd(n, k);
        n /= x;
        k /= x;
        long long y = gcd(m, k);
        m /= y;
        k /= y;
        if (k != 1)
          k /= 2;
        else if (x >= 2)
          n *= 2;
        else
          m *= 2;
        printf("%lld %lld\n%lld %lld\n", n, 0ll, 0ll, m);
      } else
        printf("%lld %lld\n%lld %lld\n", ans <= n ? ans : 0ll,
               ans > n ? ans : 0ll, ans <= n ? 0ll : 2ll, ans > n ? 0ll : 2ll);
    }
  }
  return 0;
}
