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;
map<long long, long long> prfac(long long n) {
  map<long long, long long> ans;
  if (n < 2) {
    return ans;
  }
  while (n % 2 == 0) {
    ans[2]++;
    n /= 2;
  }
  while (n % 3 == 0) {
    ans[3]++;
    n /= 3;
  }
  for (long long i = 6; (i - 1) * (i - 1) <= n; i += 6) {
    long long x = i - 1;
    while (n % x == 0) {
      ans[x]++;
      n /= x;
    }
    x = i + 1;
    while (n % x == 0) {
      ans[x]++;
      n /= x;
    }
  }
  if (n > 1) {
    ans[n]++;
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  if (n * m * 2 % k != 0) {
    cout << "NO";
    return 0;
  }
  long long lb = (2 * m + k - 1) / k;
  if (lb > m) {
    cout << "NO";
    return 0;
  }
  long long a = 2 * n * m / k;
  if (sqrt(a) < m - lb + 1) {
    for (long long i = 1; i * i <= a; i++) {
      if (a % i == 0) {
        long long x = i;
        if (x >= lb && x <= m) {
          cout << "YES\n0 0\n0 " << x << "\n" << a / x << " 0";
          return 0;
        }
        x = a / i;
        if (x >= lb && x <= m) {
          cout << "YES\n0 0\n0 " << x << "\n" << a / x << " 0";
          return 0;
        }
      }
    }
  } else {
    for (long long i = lb; i <= m; i++) {
      if (a % i == 0) {
        cout << "YES\n0 0\n" << a / i << " 0\n0 " << i;
        return 0;
      }
    }
  }
  cout << "NO";
  return 0;
}
