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();
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
}
vector<long long> divs(long long n) {
  vector<long long> ans;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      ans.push_back(i);
      ans.push_back(n / i);
    }
    if (i * i == n) ans.pop_back();
  }
  return ans;
}
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  n *= 2;
  m *= 2;
  if ((n * m) % k != 0) {
    cout << "NO" << '\n';
    return;
  }
  vector<long long> a = divs(n);
  vector<long long> b = divs(m);
  k = n * m / k;
  k *= 2;
  vector<long long> kek;
  for (auto i : a) {
    for (auto j : b) {
      if (k / i < j) break;
      if (k % (i * j) == 0) {
        kek.push_back(i * j);
      }
      if (k % (i * j * 2) == 0) {
        kek.push_back(i * j * 2);
      }
    }
  }
  sort((kek).begin(), (kek).end());
  kek.resize(unique((kek).begin(), (kek).end()) - kek.begin());
  for (auto i : kek) {
    long long n1 = i, m1 = k / i;
    if (n1 <= n && m1 <= m && n1 % 2 == 0 && m1 % 2 == 0) {
      cout << "YES" << '\n';
      cout << 0 << ' ' << 0 << '\n';
      cout << n1 / 2 << ' ' << 0 << '\n';
      cout << 0 << ' ' << m1 / 2 << '\n';
      return;
    }
  }
  cout << "NO" << '\n';
}
