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;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
long long n, m, kk;
long long gcd(long long a, long long b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  cin >> n >> m >> kk;
  if ((2 * n * m) % kk) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  printf("0 0\n");
  long long d = gcd(2 * n, kk);
  long long x = 2 * n / d;
  if (x > n) x /= 2, m *= 2;
  long long y = m * d / kk;
  printf("%lld 0\n", x);
  printf("0 %lld\n", y);
  return 0;
}
