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;
vector<long long> d1, d2;
int main() {
  scanf("%lld%lld%lld", &n, &m, &K);
  long long t = (n * m * 2) % K;
  if (t == 0) {
    long long M = n * m * 2 / K;
    for (long long i = 1; i * i <= n; i++) {
      if (n % i == 0) d1.push_back(i), d1.push_back(n / i);
    }
    for (long long i = 1; i * i <= m; i++) {
      if (m % i == 0) d2.push_back(i), d2.push_back(m / i);
    }
    long long Mx = 0;
    for (auto &x : d1)
      for (auto &y : d2) {
        long long t = x * y;
        if (t <= n && M % t == 0) Mx = max(Mx, t);
        t *= 2;
        if (t <= n && M % t == 0) Mx = max(Mx, t);
      }
    if (Mx == 0 || M / Mx > m)
      printf("NO\n");
    else {
      printf("YES\n");
      printf("%lld %lld\n", 0ll, 0ll);
      printf("%lld %lld\n", Mx, 0ll);
      printf("%lld %lld\n", 0ll, M / Mx);
    }
    return 0;
  }
  puts("NO");
}
