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(int argc, const char* argv[]) {
  int64_t n, m, k;
  cin >> n >> m >> k;
  map<int64_t, int64_t> storage_n, storage_m;
  int64_t a = n, b = m;
  for (int i = 2; i * i <= a; ++i) {
    while (a % i == 0) {
      a /= i;
      ++storage_n[i];
    }
  }
  if (a != 1) ++storage_n[a];
  a = b;
  for (int i = 2; i * i <= a; ++i) {
    while (a % i == 0) {
      a /= i;
      ++storage_m[i];
    }
  }
  if (a != 1) ++storage_m[a];
  vector<int64_t> divide_by_k;
  a = k;
  bool tmp = 1;
  if (a % 2 == 0) {
    a /= 2;
    tmp = 0;
  }
  for (int i = 2; i * i <= a; ++i) {
    while (a % i == 0) {
      a /= i;
      divide_by_k.push_back(i);
    }
  }
  if (a != 1) divide_by_k.push_back(a);
  a = n;
  b = m;
  for (auto i : divide_by_k) {
    if (storage_n[i]) {
      --storage_n[i];
      a /= i;
    } else if (storage_m[i]) {
      --storage_m[i];
      b /= i;
    } else {
      cout << "NO";
      return 0;
    }
  }
  if (tmp) {
    if (a < n)
      a *= 2;
    else
      b *= 2;
  }
  cout << "YES\n"
       << "0 " << b << "\n"
       << a << " 0\n0 0";
  return 0;
}
