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 dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
bool ckmin(int& a, int b) { return b < a ? a = b, true : false; }
bool ckmax(int& a, int b) { return b > a ? a = b, true : false; }
const long long linf = 4000000000000000000LL;
const long long inf = 1000000007;
const long double pi = 3.1415926535;
void pv(vector<int> a) {
  for (auto& x : a) cout << x << " ";
  cout << '\n';
}
void pv(vector<long long> a) {
  for (auto& x : a) cout << x << " ";
  cout << '\n';
}
void pv(vector<vector<int> > a) {
  for (int i = (0); i < (int(a.size())); ++i) {
    cout << i << '\n';
    pv(a[i]);
    cout << '\n';
  }
}
void pv(vector<vector<long long> > a) {
  for (int i = (0); i < (int(a.size())); ++i) {
    cout << i << '\n';
    pv(a[i]);
  }
  cout << '\n';
}
void pv(vector<string> a) {
  for (auto& x : a) cout << x << '\n';
  cout << '\n';
}
void setIO(string second) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (int(second.size())) {
    freopen((second + ".in").c_str(), "r", stdin);
    if (second != "test1") freopen((second + ".out").c_str(), "w", stdout);
  }
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(a, b % a);
}
int main() {
  setIO("");
  long long n, m, k;
  cin >> n >> m >> k;
  if ((2 * m * n) % k != 0) {
    cout << "NO" << '\n';
    return 0;
  }
  long long g = gcd(m, k);
  long long ans1 = m / g;
  long long ans2 = (2 * n * g) / k;
  if (g == k) ans1 *= 2, ans2 /= 2;
  cout << "YES" << '\n'
       << "0 0" << '\n'
       << "0 " << ans1 << '\n'
       << ans2 << " 0" << '\n';
}
