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;
using ll = long long;
ll a = 0, b = 0;
vector<pair<ll, ll>> pts;
void make(int n, int m, int k) {
  for (int i = 1; i * i <= k; i++) {
    if (k % i == 0 and 2 * n % i == 0 and m % (k / i) == 0 and 2 * n / i <= n) {
      a = 2 * n / i;
      b = m / (k / i);
      return;
    } else if (k % i == 0 and n % i == 0 and 2 * m % (k / i) == 0 and
               2 * m / (k / i) <= m) {
      a = n / i;
      b = 2 * m / (k / i);
      return;
    } else if (k % i == 0 and 2 * m % i == 0 and n % (k / i) == 0 and
               2 * m / i <= m) {
      b = 2 * m / i;
      a = n / (k / i);
      return;
    } else if (k % i == 0 and m % i == 0 and 2 * n % (k / i) == 0 and
               2 * n / (k / i) <= n) {
      b = m / i;
      a = 2 * n / (k / i);
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll n, m, k;
  cin >> n >> m >> k;
  if ((2 * n * m) % k == 0) {
    make(n, m, k);
  }
  if (a and b) {
    cout << "YES" << '\n';
    cout << "0 0" << '\n';
    cout << a << " " << 0 << '\n';
    cout << 0 << " " << b << '\n';
  } else {
    cout << "NO" << '\n';
  }
  return 0;
}
