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) { return !b ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  long long n, m, k;
  cin >> n >> m >> k;
  if ((2 * n * m) % k != 0) {
    cout << "NO" << '\n';
    return 0;
  }
  vector<pair<long long, long long>> ans;
  long long a = n;
  long long b = m;
  while (k > 1) {
    long long g = gcd(k, a);
    if (g == 1) {
      break;
    }
    a /= g;
    k /= g;
  }
  while (k > 1) {
    long long g = gcd(k, b);
    if (g == 1) {
      break;
    }
    b /= g;
    k /= g;
  }
  if (k == 1) {
    if (2 * a <= n) {
      ans.emplace_back(0, 0);
      ans.emplace_back(2 * a, 0);
      ans.emplace_back(0, b);
    } else if (2 * b <= m) {
      ans.emplace_back(0, 0);
      ans.emplace_back(a, 0);
      ans.emplace_back(0, 2 * b);
    }
  } else {
    ans.emplace_back(0, 0);
    ans.emplace_back(a, 0);
    ans.emplace_back(0, b);
  }
  if (ans.empty()) {
    cout << "NO" << '\n';
  } else {
    cout << "YES" << '\n';
    for (auto &each : ans) {
      cout << each.first << " " << each.second << '\n';
    }
  }
  return 0;
}
