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 int inf = 1 << 30;
const long long INF = 1ll << 60;
const double Inf = 1e20;
const double eps = 1e-9;
void gmax(int &a, int b) { a = (a > b ? a : b); }
void gmin(int &a, int b) { a = (a < b ? a : b); }
long long n, m, k;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
int main() {
  cin >> n >> m >> k;
  if (n * m * 2 % k) {
    cout << "NO";
    return 0;
  }
  long long x = gcd(n * 2, k), y;
  if (x != 1) {
    k /= x;
    x = n * 2 / x;
    y = m / gcd(m, k);
  } else {
    int w = gcd(m * 2, k);
    y = m * 2 / w;
    k /= w;
    x = n / gcd(n, k);
  }
  printf("YES\n0 0\n0 %lld\n%lld 0\n", y, x);
  return 0;
}
