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;
int gcd(int x, int y) { return x == 0 ? y : gcd(y % x, x); }
int main() {
  if (fopen("input.txt", "r")) freopen("input.txt", "r", stdin);
  int n, m, k, orn, orm, ork;
  scanf("%d %d %d ", &n, &m, &k);
  orn = n;
  orm = m;
  ork = k;
  int u = gcd(n, k);
  n /= u;
  k /= u;
  int v = gcd(m, k);
  m /= v;
  k /= v;
  if (k == 2) {
    printf("YES\n");
    printf("0 0\n%d 0\n0 %d", n, m);
    return 0;
  }
  if (k != 1) {
    printf("NO");
    return 0;
  }
  if (n != orn) {
    printf("YES\n");
    printf("0 0\n%d 0\n0 %d", 2 * n, m);
  } else if (m != orm) {
    printf("YES\n");
    printf("0 0\n%d 0\n0 %d", n, 2 * m);
  }
  return 0;
}
