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;
void Debug() { cout << '\n'; }
template <class FIRST, class... REST>
void Debug(FIRST arg, REST... rest) {
  cout << arg << " ";
  Debug(rest...);
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  if (!v.empty()) {
    for (int i = int(0); i < int((int)(v).size() - 1); i++) out << v[i] << ", ";
    out << v.back();
  }
  out << "]";
  return out;
}
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& v) {
  out << "(" << v.first << ", " << v.second << ")";
  return out;
}
const int MAX_N = 200010;
const int MAX_V = 100010;
const double eps = 1e-6;
const long long mod = 1000000007;
const int inf = 1 << 30;
const long long linf = 1LL << 60;
const double PI = 3.14159265358979323846;
long long N, M, K;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
void solve() {
  cin >> N >> M >> K;
  long long lcm = N * M * 2;
  if (lcm % K != 0) {
    cout << "NO\n";
  } else {
    long long v = lcm / K;
    long long x = N / gcd(N, K);
    if (x != N && (v / x) % 2 == 0) x *= 2;
    long long y = v / x;
    if (x <= N && y <= M) {
      cout << "YES\n";
      cout << 0 << " " << 0 << "\n";
      cout << x << " " << 0 << "\n";
      cout << 0 << " " << y << "\n";
    } else
      cout << "NO\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed;
  cout.precision(20);
  srand((unsigned int)time(NULL));
  solve();
  return 0;
}
