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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
inline long long pow(long long a, long long b, long long p) {
  long long rtn = 1;
  while (b) {
    if (b & 1) rtn = rtn * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return rtn;
}
long long inv(long long a, long long p) { return pow(a, p - 2, p); }
long long gcd(long long a, long long b) {
  long long k;
  while (b != 0) {
    k = b;
    b = a % b;
    a = k;
  }
  return a;
}
void makedata() {
  freopen("input.txt", "w", stdout);
  srand((unsigned)time(NULL));
  fclose(stdout);
}
int main() {
  std::ios::sync_with_stdio(0), cin.tie(0);
  long long n, m, k, p, a, b;
  cin >> n >> m >> k;
  if (2 * n * m % k) {
    cout << ("NO") << endl;
    return 0;
  }
  p = 2 * n * m / k;
  a = gcd(p, n);
  b = p / a;
  if (a <= n && b <= m) {
    cout << ("YES") << endl;
    cout << a << ' ' << 0 << endl;
    cout << 0 << ' ' << 0 << endl;
    cout << 0 << ' ' << b << endl;
  } else {
    b = gcd(p, m);
    a = p / b;
    cout << ("YES") << endl;
    cout << a << ' ' << 0 << endl;
    cout << 0 << ' ' << 0 << endl;
    cout << 0 << ' ' << b << endl;
  }
  return 0;
}
