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 maxN = 7e3 + 9, MOD = 998244353, lg = 19;
long long n, m, k;
long long orign, origm;
bool isvalid(pair<long long, long long> pai) {
  return pai.first <= orign && pai.second <= origm;
}
pair<long long, long long> getintcoor(long long x) {
  for (long long i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      long long p = i, q = x / i, a, b;
      if (n % p == 0 && m % q == 0) {
        a = n / p, b = m / q;
        if (isvalid(pair<long long, long long>(a, b)))
          return pair<long long, long long>(a, b);
      }
      if (m % p == 0 && n % q == 0) {
        a = m / p, b = n / q;
        if (isvalid(pair<long long, long long>(b, a)))
          return pair<long long, long long>(b, a);
      }
    }
  }
  return pair<long long, long long>(-1, -1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  orign = n, origm = m;
  if (n * m * 2 % k != 0) {
    cout << "NO\n";
    return 0;
  }
  n *= 2;
  pair<long long, long long> crr = getintcoor(k);
  if (crr.first != -1) {
    cout << "YES\n";
    cout << "0 0\n";
    cout << crr.first << " 0\n";
    cout << "0 " << crr.second << '\n';
    return 0;
  }
  n /= 2;
  m *= 2;
  crr = getintcoor(k);
  if (crr.first == -1) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  cout << "0 0\n";
  cout << crr.first << " 0\n";
  cout << "0 " << crr.second << '\n';
}
