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(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, m, k;
    cin >> n >> m >> k;
    if ((2 * n * m) % k != 0)
      cout << "NO" << '\n';
    else {
      map<long long, long long> m1, m2, m3;
      long long temp = n;
      for (long long i = 2; i <= sqrt(temp); i++) {
        if (temp % i == 0) {
          while (temp % i == 0) {
            temp /= i;
            m1[i]++;
          }
        }
      }
      if (temp > 1) m1[temp]++;
      temp = m;
      for (long long i = 2; i <= sqrt(temp); i++) {
        if (temp % i == 0) {
          while (temp % i == 0) {
            temp /= i;
            m2[i]++;
          }
        }
      }
      if (temp > 1) m2[temp]++;
      temp = k;
      for (long long i = 2; i <= sqrt(temp); i++) {
        if (temp % i == 0) {
          while (temp % i == 0) {
            temp /= i;
            m3[i]++;
          }
        }
      }
      if (temp > 1) m3[temp]++;
      bool flag = false;
      if (m3[2] > 0) {
        m3[2]--;
        flag = true;
      }
      long long x = n;
      long long y = m;
      for (auto it = m3.begin(); it != m3.end(); it++) {
        while (it->second > 0) {
          it->second--;
          if (x % it->first == 0)
            x /= it->first;
          else
            y /= it->first;
        }
      }
      if (flag == false) {
        if (2 * x <= n)
          x *= 2;
        else
          y *= 2;
      }
      cout << "YES" << '\n';
      cout << 0 << " " << 0 << '\n';
      cout << x << " " << 0 << '\n';
      cout << 0 << " " << y << '\n';
    }
  }
  return 0;
}
