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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  long long sn = n, sm = m;
  if ((n * m * 2) % k != 0) {
    cout << "NO\n";
    return 0;
  }
  long long toGet = k;
  bool twoNeeded = true;
  if (toGet % 2 == 0) {
    twoNeeded = false;
    toGet /= 2;
  }
  long long prime = 2;
  while (toGet != 1) {
    if (toGet % prime != 0) {
      prime++;
      continue;
    }
    while (n % prime == 0 && toGet % prime == 0) {
      n /= prime;
      toGet /= prime;
    }
    while (m % prime == 0 && toGet % prime == 0) {
      m /= prime;
      toGet /= prime;
    }
    if (n % toGet == 0) {
      n /= toGet;
      toGet = 1;
    }
    if (m % toGet == 0) {
      m /= toGet;
      toGet = 1;
    }
  }
  if (twoNeeded) {
    if (n * 2 <= sn)
      n *= 2;
    else
      m *= 2;
  }
  cout << "YES\n0 0\n" << n << " 0\n0 " << m << "\n";
  return 0;
}
