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 = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
long long read_long() {
  long long ans = 0;
  char c;
  bool started = false;
  bool is_negative = false;
  while (true) {
    c = getchar();
    if ((c < '0' || c > '9') && c != '-' && !started) continue;
    if ((c < '0' || c > '9') && c != '-' && started) break;
    if (started) {
      ans = ans * 10;
    }
    started = true;
    if (c == '-')
      is_negative = true;
    else
      ans += c - '0';
  }
  if (is_negative) ans *= -1;
  return ans;
}
int n, m, k;
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return gcd(b, a % b);
}
int main() {
  cin >> n >> m >> k;
  long long p = n * 1ll * m;
  long long q = k;
  long long d = gcd(p, q);
  p /= d;
  q /= d;
  if (q > 2) {
    printf("NO\n");
    return 0;
  }
  long long d1 = gcd(n, k);
  long long n1 = n / d1;
  long long k1 = k / d1;
  long long d2 = gcd(m, k1);
  long long m1 = m / d2;
  long long k2 = k1 / d2;
  if (k2 == 2)
    printf("YES\n0 0\n%lld 0\n0 %lld\n", n1, m1);
  else if (n1 * 2ll <= n)
    printf("YES\n0 0\n%lld 0\n0 %lld\n", 2ll * n1, m1);
  else
    printf("YES\n0 0\n%lld 0\n0 %lld\n", n1, m1 * 2ll);
  return 0;
}
