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;
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long in = n, im = m;
  if ((2 * n * m) % k != 0) {
    cout << "NO\n";
    return;
  }
  bool poss = true;
  if (k % 2 == 0)
    k /= 2;
  else
    poss = false;
  vector<long long> fac;
  fac.push_back(k);
  for (long long i = 2; i * i <= k; i++) {
    if (k % i == 0) {
      fac.push_back(i);
      fac.push_back(k / i);
    }
  }
  sort(fac.begin(), fac.end());
  for (long long i = fac.size() - 1; i >= 0; i--) {
    if (k == 1) break;
    if (n % fac[i] == 0 && k % fac[i] == 0) {
      n /= fac[i];
      k /= fac[i];
    }
    if (m % fac[i] == 0 && k % fac[i] == 0) {
      m /= fac[i];
      k /= fac[i];
    }
  }
  if (!poss) {
    if (n * 2 <= in)
      n *= 2, poss = true;
    else if (m * 2 <= im)
      m *= 2, poss = true;
  }
  if (!poss) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  cout << "0 0" << endl;
  cout << "0 " << m << endl;
  cout << n << " 0" << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
}
