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 ll = long long;
void dump(const vector<int>& a) {
  for (int i = 0; i < a.size(); i++) {
    cerr << a[i] << ' ';
  }
  cerr << endl;
}
vector<pair<int, int>> factorise(long long m) {
  vector<pair<int, int>> res;
  for (long long i = 2; i * i <= m; i++) {
    if (m % i == 0) {
      int cnt = 0;
      while (m % i == 0) {
        cnt++;
        m /= i;
      }
      res.emplace_back(i, cnt);
    }
  }
  return res;
};
using ll = long long;
ll gcd(ll a, ll b, ll& x, ll& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  ll x1, y1;
  ll d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
bool debug = false;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
const int magic = 3;
const int INF = 1e9;
void solve();
void init();
int main(int argc, char** argv) {
  cout.precision(15);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  for (int tt = 1; tt <= t; tt++) {
    solve();
  }
  return 0;
}
const int MX = 2100500;
int gcd(int x, int y) {
  while (y > 0) {
    x %= y;
    swap(x, y);
  }
  return x;
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  int g = gcd(n, k);
  n /= g;
  k /= g;
  bool x = g > 1;
  g = gcd(m, k);
  m /= g;
  k /= g;
  if (k > 2) {
    cout << "NO\n";
    return;
  }
  if (k == 1) {
    if (x) {
      n *= 2;
    } else {
      m *= 2;
    }
  }
  cout << "YES\n";
  cout << "0 0\n";
  cout << n << " 0\n";
  cout << "0 " << m << endl;
}
