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 = 1000000000 + 7;
const long long LL_INF = 1ll * INF * INF;
const int MAX_N = 100000 + 7;
template <typename T>
inline void addmod(T& a, const long long& b, const int& MOD = INF) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
long long m, n, k;
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
void solve() {
  cin >> n >> m >> k;
  if (2 * m * n % k != 0) {
    cout << "NO";
  } else {
    cout << "YES\n";
    cout << 0 << ' ' << 0 << '\n';
    long long tmp = 2 * m / gcd(2 * m, k);
    if (tmp <= m) {
      cout << 0 << ' ' << 2 * m / gcd(2 * m, k) << '\n';
      k /= gcd(2 * m, k);
      cout << n / k << ' ' << 0 << '\n';
    } else {
      cout << 0 << ' ' << m / gcd(m, k) << '\n';
      k /= gcd(m, k);
      cout << 2 * n / k << ' ' << 0 << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}
