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;
inline bool isvowel(char c) {
  c = tolower(c);
  if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u')
    return 1;
  return 0;
}
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 6e5 + 123;
const int M = 22;
const int pri = 997;
const int Magic = 2101;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
long long n, m, k;
int cnt[N];
string s;
int a[N];
vector<int> g[N];
int u[N];
int ans, res;
int cur;
long long other;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m >> k;
  if ((2 * n * m) % k == 0) {
    long long now = 2 * n * m;
    now /= k;
    other = now;
    if (max(n, m) < other) other = max(n, m);
    if (now % n == 0 && now / n <= m) {
      cout << "YES\n0 0\n" << n << " 0\n0 " << now / n;
      exit(0);
    }
    if (now % m == 0 && now / m <= n) {
      cout << "YES\n0 0\n" << now / m << " 0\n0 " << m;
      exit(0);
    }
    for (long long i = 1; i <= other; i++) {
      if (i * i > now) break;
      if (now % i) continue;
      if (i <= n && now / i <= m) {
        cout << "YES\n0 0\n" << i << " 0\n0 " << now / i;
        exit(0);
      }
      if (i <= m && now / i <= n) {
        cout << "YES\n0 0\n" << now / i << " 0\n0 " << i;
        exit(0);
      }
    }
  }
  cout << "NO";
  return 0;
}
