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;
template <typename tp>
void read(tp& dig) {
  char ch = getchar();
  dig = 0;
  long long flag = 0;
  while (!isdigit(ch)) {
    if (ch == '-') flag = 1;
    ch = getchar();
  }
  while (isdigit(ch)) dig = dig * 10 + ch - '0', ch = getchar();
  if (flag) dig = -dig;
}
long long n, m, k, K, flag;
int main() {
  read(n);
  read(m);
  read(k);
  while (!(k & 1)) k >>= 1, ++K;
  if (k == 1) {
    --K;
    while (K && !(n & 1)) --K, n >>= 1;
    while (K && !(m & 1)) --K, m >>= 1;
    if (K) {
      printf("NO\n");
      return 0;
    }
    printf("YES\n");
    printf("0 0\n");
    printf("%lld 0\n", n);
    printf("0 %lld\n", m);
    return 0;
  }
  if (n < m) swap(n, m), flag = 1;
  if (n * m % k) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  long long area = n * m / k * 2, q = sqrt(area), x, y;
  for (long long i = max(1LL, area / n - 1); i; ++i)
    if (area % i == 0) {
      if (area / i > n) continue;
      y = i, x = area / i;
      break;
    }
  if (x < y) swap(x, y);
  if (flag) swap(x, y);
  while (K && !(x & 1)) --K, x >>= 1;
  while (K && !(y & 1)) --K, y >>= 1;
  printf("0 0\n");
  printf("%lld 0\n", x);
  printf("0 %lld\n", y);
  return 0;
}
