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 long long read() {
  long long x = 0, f = 1;
  char cc = getchar();
  while (cc < '0' || cc > '9') {
    if (cc == '-') f = -1;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') {
    x = x * 10 + cc - '0';
    cc = getchar();
  }
  return x * f;
}
long long n, m, k, s, g, a, b;
inline long long gcd(long long a, long long b) {
  while (b) {
    long long r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  n = read();
  m = read();
  k = read();
  s = n * m;
  g = gcd(s, k);
  if (k / g > 2) {
    printf("NO");
    return 0;
  }
  a = gcd(n, k);
  k /= a;
  a = n / a;
  b = gcd(m, k);
  k /= b;
  b = m / b;
  printf("YES\n");
  printf("0 0\n");
  if (k == 2) {
    printf("%I64d 0\n0 %I64d", a, b);
    return 0;
  }
  if (a * 2 <= n)
    a *= 2;
  else
    b *= 2;
  printf("%I64d 0\n0 %I64d", a, b);
}
