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;
long long n, m, k;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  scanf("%lld %lld %lld", &n, &m, &k);
  if (k / gcd(n * m, k) == 1) {
    long long base, h, tmp = gcd(n, k);
    base = n / tmp;
    h = m * tmp / k;
    if (2 * base <= n)
      base *= 2;
    else
      h *= 2;
    printf("YES\n");
    printf("0 0\n");
    printf("%lld 0\n", base);
    printf("0 %lld\n", h);
  } else if (k / gcd(n * m, k) == 2) {
    printf("Yes\n");
    long long tmp = gcd(k, n) * gcd(k, m) / k;
    long long base, h;
    long long tmp2 = k;
    base = n / gcd(tmp2, n);
    tmp2 /= gcd(tmp2, n);
    h = m * 2 / tmp2;
    printf("0 0\n");
    printf("%lld 0\n", base);
    printf("0 %lld\n", h);
  } else {
    printf("NO\n");
  }
}
