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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
lli gcd(lli a, lli b) { return (b == 0) ? a : gcd(b, a % b); }
int tejas_919(int kkkk) {
  lli n, m, k, q, u, v, temp = 0, ans = 0;
  cin >> n >> m >> k;
  temp = gcd(n * m, k);
  k /= temp;
  if (k > 2) return cout << "NO" << '\n', 0;
  cout << "YES" << '\n';
  if (k == 1) {
    u = gcd(n, temp);
    bool fl = 1;
    if (u >= 2) {
      n /= u;
      temp /= u;
      n *= 2;
      fl = 0;
    }
    u = gcd(m, temp);
    if (u >= 2) {
      m /= u;
      temp /= u;
      if (fl) m *= 2;
    }
    cout << 0 << " " << 0 << '\n';
    cout << 0 << " " << m << '\n';
    cout << n << " " << 0 << '\n';
  } else {
    u = gcd(n, temp);
    n /= u;
    temp /= u;
    u = gcd(m, temp);
    m /= u;
    temp /= u;
    cout << 0 << " " << 0 << '\n';
    cout << 0 << " " << m << '\n';
    cout << n << " " << 0 << '\n';
  }
  return 0;
}
signed main() {
  if (!0) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    tejas_919(i + 1);
  }
  {};
  if (0) system("pause");
}
