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;
const int inf = 0x3f3f3f3f;
const int maxn = 2e3 + 10;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  long long n, m, k;
  scanf("%lld %lld %lld", &n, &m, &k);
  if (2ll * n * m % k != 0) {
    puts("NO");
    return 0;
  }
  puts("YES");
  puts("0 0");
  long long x2, y2, x3, y3;
  long long sum = 2ll * n * m / k;
  if (gcd(2ll * n, k) == 1) {
    long long u = gcd(2ll * m, k);
    printf("%lld %lld\n%lld %lld", n / (k / u), 0ll, 0ll, 2ll * m / u);
  } else {
    long long u = gcd(2ll * n, k);
    printf("%lld %lld\n%lld %lld", 2ll * n / u, 0ll, 0ll, m / (k / u));
  }
  return 0;
}
