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;
int read() {
  char ch = getchar();
  int f = 0, x = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    f = (f << 1) + (f << 3) + ch - '0';
    ch = getchar();
  }
  return f * x;
}
long long n, m, k;
int main() {
  cin >> n >> m >> k;
  if (k % 2 == 0) {
    k /= 2;
    int lim = sqrt(k);
    for (int i = 2; i <= lim; i++) {
      while (k % i == 0) {
        if (n % i == 0)
          n /= i;
        else if (m % i == 0)
          m /= i;
        else {
          puts("NO");
          return 0;
        }
        k /= i;
      }
    }
    if (n % k == 0)
      n /= k;
    else if (m % k == 0)
      m /= k;
    else {
      puts("NO");
      return 0;
    }
    puts("YES");
    puts("0 0");
    cout << n << " " << 0 << endl;
    cout << 0 << " " << m << endl;
  } else {
    int lim = sqrt(k);
    int n1 = n, m1 = m;
    for (int i = 2; i <= lim; i++) {
      while (k % i == 0) {
        if (n % i == 0)
          n /= i;
        else if (m % i == 0)
          m /= i;
        else {
          puts("NO");
          return 0;
        }
        k /= i;
      }
    }
    if (n % k == 0)
      n /= k;
    else if (m % k == 0)
      m /= k;
    else {
      puts("NO");
      return 0;
    }
    if (n * 2 <= n1)
      n *= 2;
    else if (m * 2 <= m1)
      m *= 2;
    puts("YES");
    puts("0 0");
    cout << n << " " << 0 << endl;
    cout << 0 << " " << m << endl;
  }
}
