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 GCC optimize("-O3")
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void Out(int a) {
  if (a > 9) Out(a / 10);
  putchar(a % 10 + '0');
}
const int inf = 1e9;
const long long mod = 1e9 + 7;
long long n, m, k, S;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  cin >> n >> m >> k;
  long long a, N = n, M = m;
  a = gcd(k, n);
  n /= a;
  k /= a;
  a = gcd(k, m);
  m /= a;
  k /= a;
  if (k == 2) {
    printf("YES\n");
    cout << 0 << " " << 0 << endl;
    cout << 0 << " " << m << endl;
    cout << n << " " << 0 << endl;
    return 0;
  } else if (k == 1) {
    if (n + n <= N) {
      n = n + n;
      printf("YES\n");
      cout << 0 << " " << 0 << endl;
      cout << 0 << " " << m << endl;
      cout << n << " " << 0 << endl;
      return 0;
    }
    if (m + m <= M) {
      m = m + m;
      printf("YES\n");
      cout << 0 << " " << 0 << endl;
      cout << 0 << " " << m << endl;
      cout << n << " " << 0 << endl;
      return 0;
    }
    printf("NO\n");
    return 0;
  } else {
    printf("NO\n");
    return 0;
  }
}
