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;
template <typename T>
inline void setmin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
inline void setmax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
const int MAX = 2e5 + 30;
const int INF = 1e9 + 7;
const long long BIG_INF = 1e18 + 5;
long long n, m, k;
long long pole;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  long long pom_n = n;
  long long pom_m = m;
  long long temp = gcd(n, k);
  n /= temp;
  k /= temp;
  temp = gcd(m, k);
  m /= temp;
  k /= temp;
  if (k > 2) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  if (k == 1) {
    if (m * 2 <= pom_m) {
      cout << "0 0\n";
      cout << n << ' ' << 0 << '\n';
      cout << 0 << ' ' << m * 2 << '\n';
    } else {
      cout << "0 0\n";
      cout << n * 2 << ' ' << 0 << '\n';
      cout << 0 << ' ' << m << '\n';
    }
  } else {
    cout << "0 0\n";
    cout << n << ' ' << 0 << '\n';
    cout << 0 << ' ' << m << '\n';
  }
}
