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 = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const long double pi = acos(-1);
const int MOD = 1e9 + 7;
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long area = 2 * (n * m);
  if (area % k) {
    cout << "NO" << endl;
    return 0;
  }
  area /= k;
  vector<long long> divi;
  for (long long i = 1; i * i <= k; i++) {
    if (k % i == 0) {
      divi.push_back(i);
      divi.push_back(k / i);
    }
  }
  for (long long x : divi) {
    long long tirax = x;
    long long tiray = k / x;
    long long X = (2 * n) / tirax;
    long long Y = m / tiray;
    if (tirax == 3) {
      cerr << "X"
           << " == " << X << endl;
      ;
      cerr << "Y"
           << " == " << Y << endl;
      ;
    }
    if (X * Y == area and X <= n and Y <= m) {
      cout << "YES" << endl;
      cout << 0 << " " << 0 << endl;
      cout << X << " " << 0 << endl;
      cout << 0 << " " << Y << endl;
      return 0;
    }
  }
  for (long long x : divi) {
    long long tirax = x;
    long long tiray = k / x;
    long long X = n / tirax;
    long long Y = (2 * m) / tiray;
    if (X * Y == area and X <= n and Y <= m) {
      cout << "YES" << endl;
      cout << 0 << " " << 0 << endl;
      cout << X << " " << 0 << endl;
      cout << 0 << " " << Y << endl;
      return 0;
    }
  }
  assert(false);
}
