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 N = 1e6 + 229;
const int M = 1e9 + 7;
const int rx[4] = {1, -1, 0, 0};
const int ry[4] = {0, 0, 1, -1};
const double eps = 0.000001;
void unite_set(int a, int b, long long x) {}
long long bp(long long a, long long n) {
  long long otved = 1;
  while (n) {
    if (n & 1) otved *= a;
    otved %= M;
    a *= a;
    a %= M;
    n >>= 1;
  }
  return otved;
}
int find_set(int v) {}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long double n, m, k;
  cin >> n >> m >> k;
  long double second = 2 * n * m / k;
  long double dif = abs(second - (long long)second);
  if (dif != 0 && dif != 0.5) {
    cout << "NO";
    return 0;
  } else {
    set<long long> p;
    set<long long> pryvet;
    long long otved = second;
    long long nyancat = n;
    long long memmi = m;
    for (long long i = 1; 1ll * i * i <= nyancat; i++) {
      if (nyancat % i == 0) {
        p.insert(i);
        p.insert(nyancat / i);
      }
    }
    for (long long i = 1; 1ll * i * i <= memmi; i++) {
      if (memmi % i == 0) {
        pryvet.insert(i);
        pryvet.insert(memmi / i);
      }
    }
    set<long long> st;
    for (auto x : p) {
      for (auto y : pryvet) {
        if (1ll * x * y <= n || 1ll * x * y <= m) st.insert(1ll * x * y);
        st.insert(x);
        st.insert(y);
      }
    }
    for (auto x : st) {
      long long y = otved / x;
      long double s = (long double)x * (long double)y;
      if (abs(second - s) < 1e-9) {
        if (x <= n && y <= m) {
          cout << "YES" << '\n';
          cout << "0 0" << '\n';
          cout << x << " 0" << '\n';
          cout << "0 " << y << '\n';
          return 0;
        }
        if (y <= n && x <= m) {
          cout << "YES" << '\n';
          cout << "0 0" << '\n';
          cout << y << " 0" << '\n';
          cout << "0 " << x << '\n';
          return 0;
        }
      }
    }
    cout << "NO";
  }
  return 0;
}
