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;
long long n, m, k;
vector<long long> divisors(long long x) {
  vector<long long> res;
  for (long long i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      res.push_back(i);
      res.push_back(x / i);
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> k;
  if (k == 2) {
    cout << "YES\n";
    cout << "0 0\n";
    cout << "0 " << m << endl;
    cout << n << " 0" << endl;
  } else if ((n * m) % k == 0 && k > 1) {
    vector<long long> d = divisors(k);
    long long curd = 1;
    for (int i = 0; i < d.size(); i++) {
      if (n % d[i] == 0 && m % (k / d[i]) == 0) {
        curd = max(curd, d[i]);
      }
    }
    n /= curd;
    m /= (k / curd);
    if (curd == 1) {
      cout << "YES\n";
      cout << "0 0\n";
      cout << "0 " << m * 2 << endl;
      cout << n << " 0" << endl;
    } else {
      cout << "YES\n";
      cout << "0 0\n";
      cout << "0 " << m << endl;
      cout << 2 * n << " 0" << endl;
    }
  } else if (k % 2 == 0 && (n * m) % (k / 2) == 0) {
    k /= 2;
    vector<long long> d = divisors(k);
    long long curd = 1;
    for (int i = 0; i < d.size(); i++) {
      if (n % d[i] == 0 && m % (k / d[i]) == 0) {
        curd = max(curd, d[i]);
      }
    }
    n /= curd;
    m /= (k / curd);
    if (curd == 1) {
      cout << "YES\n";
      cout << "0 0\n";
      cout << "0 " << m << endl;
      cout << n << " 0" << endl;
    } else {
      cout << "YES\n";
      cout << "0 0\n";
      cout << "0 " << m << endl;
      cout << n << " 0" << endl;
    }
  } else {
    cout << "NO\n";
  }
  return 0;
}
