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;
const int inv = 1000000000;
const int minv = -inv;
long long fgcd(long long a, long long b) {
  if (a == 0ll)
    return b;
  else
    return fgcd(b % a, a);
}
long long m, n, k;
int main() {
  cin >> m;
  cin >> n;
  cin >> k;
  if ((2ll * m * n) % k != 0ll) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  if (k & 1ll) {
    long long x = fgcd(m, k);
    long long y = k / x;
    if (x > 1ll) {
      cout << 0ll << " ";
      cout << 0ll << "\n";
      cout << (2ll * m) / x << " ";
      cout << 0ll << "\n";
      cout << 0ll << " ";
      cout << n / y << "\n";
    } else {
      cout << 0ll << " ";
      cout << 0ll << "\n";
      cout << m / x << " ";
      cout << 0ll << "\n";
      cout << 0ll << " ";
      cout << (2ll * n) / y << "\n";
    }
  } else {
    k = k / 2ll;
    long long x = fgcd(m, k);
    long long y = k / x;
    cout << 0ll << " ";
    cout << 0ll << "\n";
    cout << m / x << " ";
    cout << 0ll << "\n";
    cout << 0ll << " ";
    cout << n / y << "\n";
  }
  return 0;
}
