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>
const int maxn = 0x3f3f3f3f;
const double EI =
    2.71828182845904523536028747135266249775724709369995957496696762772407663035354594571382178525166427;
const double PI = 3.141592653589793238462643383279;
using namespace std;
inline long long gcd(long long n, long long m) {
  if (n < m) {
    swap(n, m);
  }
  long long r;
  while (n % m) {
    r = n % m;
    n = m;
    m = r;
  }
  return m;
}
int main(void) {
  long long n, m, k, n1, m1, x, y, s;
  while (~scanf("%lld %lld %lld", &n, &m, &k)) {
    s = n * m * 2 / k;
    if (n * m * 2 % k) {
      printf("NO\n");
      continue;
    } else {
      n1 = n;
      m1 = m;
      n *= 2;
      x = gcd(n, k);
      y = gcd(m, k);
      if (x >= 2) {
        n /= x;
        m = s / n;
      } else {
        m /= y;
        n = s / m;
        n /= 2;
        m *= 2;
      }
      printf("YES\n");
      printf("0 0\n");
      printf("%lld 0\n", n);
      printf("0 %lld\n", m);
    }
  }
  return 0;
}
