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 x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  scanf("%I64d%I64d%I64d", &N, &M, &K);
  long long g = gcd(N * M, K);
  long long x = N * M / g, y = K / g;
  if (y > 2) {
    puts("NO");
    return 0;
  }
  long long t = gcd(2 * N, K);
  if (t >= 2) {
    long long a = 2 * N / t;
    long long b = 2 * N * M / a / K;
    printf("YES\n");
    printf("0 0\n");
    printf("0 %I64d\n", b);
    printf("%I64d 0\n", a);
  } else {
    t = gcd(2 * M, K);
    long long b = 2 * M / t;
    long long a = 2 * M * N / b / K;
    printf("YES\n");
    printf("0 0\n");
    printf("0 %I64d\n", b);
    printf("%I64d 0\n", a);
  }
  return 0;
}
