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>
#pragma GCC optimize("Ofast")
using namespace std;
void solve();
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(2281337);
  solve();
  return 0;
}
const double eps = 1e-9;
const int maxn = (int)1e2 + 7;
const int mod = (int)1e9 + 7;
const int inf = (int)1e9;
long long gcd(long long a, long long b) {
  while (a != 0 && b != 0) {
    if (b > a) swap(a, b);
    a %= b;
  }
  return b;
}
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  if ((n * m) % k != 0 && k / gcd(n * m, k) != 2) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  if (k / gcd(n * m, k) == 2) {
    k /= 2;
    long long g = gcd(n, k), g1 = k / g;
    cout << 0 << " " << 0 << "\n";
    cout << n / g << " " << 0 << "\n";
    cout << 0 << " " << m / g1 << "\n";
  } else if (n % k == 0) {
    cout << 0 << " " << 0 << "\n";
    cout << n / k * 2ll << " " << 0 << "\n";
    cout << 0 << " " << m << "\n";
  } else if (m % k == 0) {
    cout << 0 << " " << 0 << "\n";
    cout << n << " " << 0 << "\n";
    cout << 0 << " " << m / k * 2ll << "\n";
  } else {
    long long g = gcd(n, k), g1 = k / g;
    cout << 0 << " " << 0 << "\n";
    cout << n / g * 2ll << " " << 0 << "\n";
    cout << 0 << " " << m / g1 << "\n";
  }
  return;
}
