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 gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
bool ok(long long a, long long d, long long m) {
  if (a % d) return false;
  return (a / d) <= m;
}
void print(long long d, long long e) {
  cout << "YES" << endl;
  cout << "0 0" << endl;
  cout << d << " " << 0 << endl;
  cout << 0 << " " << e << endl;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  long long n, m, k;
  cin >> n >> m >> k;
  long long a = n * m;
  long long g = gcd(a, k);
  a /= g;
  k /= g;
  if (k > 2) {
    cout << "NO" << endl;
    return 0;
  }
  if (k == 1) a *= 2;
  for (long long d = 1; d * d <= n; d++) {
    if (n % d == 0) {
      long long e = n / d;
      if (ok(a, d, m)) {
        print(d, a / d);
        return 0;
      }
      if (ok(a, e, m)) {
        print(e, a / e);
        return 0;
      }
    }
  }
  for (long long e = 1; e * e <= m; e++) {
    if (m % e == 0) {
      long long d = m / e;
      if (ok(a, e, n)) {
        print(a / e, e);
        return 0;
      }
      if (ok(a, d, n)) {
        print(a / d, d);
        return 0;
      }
    }
  }
  cout << "NO" << endl;
  return 0;
}
