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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long INF = 1000LL * 1000 * 1000 * 1000 * 1000 * 1000;
const int inf = 1000 * 1000 * 1000;
const long double PI = acos(-1.0);
const long long mod1 = inf + 7;
const long long mod2 = inf + 9;
const int MAXN = 1000005;
const long double EPS = 1e-11;
int hp = 179;
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  ;
  long long n, m, k;
  cin >> n >> m >> k;
  if ((long double)n * m / k + EPS < 0.5) {
    cout << "NO";
    return 0;
  }
  if ((n * m * 2) % k != 0) {
    cout << "NO";
    return 0;
  }
  if (k == 2) {
    cout << "YES" << endl;
    cout << "0 0\n" << n << ' ' << 0 << '\n' << 0 << ' ' << m;
    return 0;
  }
  if (k % 2) {
    long long sq = 2 * n * m / k;
    for (long long i = 1; i * i <= sq; ++i) {
      if ((sq) % i != 0) {
        continue;
      }
      if (i <= n && (sq / i) <= m) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << i << ' ' << 0 << endl;
        cout << 0 << ' ' << sq / i << endl;
        return 0;
      }
    }
    for (long long i = 1; i * i <= sq; ++i) {
      if (sq % i != 0) {
        continue;
      }
      if (i <= m && sq / i <= n) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << sq / i << ' ' << 0 << endl;
        cout << 0 << ' ' << i << endl;
        return 0;
      }
    }
    cout << "NO";
    return 0;
  } else {
    k /= 2;
    if (n % k == 0) {
      cout << "YES" << endl;
      cout << "0 0" << endl;
      cout << n / k << " " << 0 << endl;
      cout << 0 << ' ' << m;
      return 0;
    } else if (m % k == 0) {
      cout << "YES" << endl;
      cout << "0 0" << endl;
      cout << n << " " << 0 << endl;
      cout << 0 << ' ' << m / k;
      return 0;
    }
    long long sq = n * m / k;
    for (long long i = 1; i * i <= sq; ++i) {
      if ((sq) % i != 0) {
        continue;
      }
      if (i <= n && (sq / i) <= m) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << i << ' ' << 0 << endl;
        cout << 0 << ' ' << sq / i << endl;
        return 0;
      }
    }
    for (long long i = 1; i * i <= sq; ++i) {
      if (sq % i != 0) {
        continue;
      }
      if (i <= m && sq / i <= n) {
        cout << "YES" << endl;
        cout << "0 0" << endl;
        cout << sq / i << ' ' << 0 << endl;
        cout << 0 << ' ' << i << endl;
        return 0;
      }
    }
    cout << "NO";
  }
}
