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;
vector<long long> dvsn, dvsm, dvs, res;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  if (n * m * 2 % k != 0) return cout << "NO", 0;
  long long cur = n * m * 2 / k;
  long long temp = 2 * n;
  if (temp % k == 0)
    temp = temp / k;
  else
    temp = temp / k + 1;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      dvsn.push_back(i);
      if (i != n / i) dvsn.push_back(n / i);
    }
  }
  for (int i = 1; i <= sqrt(m); i++) {
    if (m % i == 0) {
      dvsm.push_back(i);
      if (i != m / i) dvsm.push_back(m / i);
    }
  }
  for (auto i : dvsn) {
    for (auto j : dvsm) {
      dvs.push_back(i * j);
      dvs.push_back(2 * i * j);
    }
  }
  dvs.resize(unique(dvs.begin(), dvs.end()) - dvs.begin());
  for (auto i : dvs)
    if (i % k == 0) res.push_back(i / k);
  for (auto i : res) {
    if (temp <= i && i <= n) {
      cout << "YES\n0 0\n";
      cout << i << " " << 0 << "\n";
      cout << 0 << " " << cur / i;
      return 0;
    }
  }
  cout << "NO";
  return 0;
}
