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;
struct P {
  int x, y;
  bool operator<(const P &a) const { return y > a.y; }
};
vector<int> v;
int a, c, i, b, n, m, k, d;
int o[1];
int l[1];
int j[1];
int e;
int dx[10] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[10] = {1, 0, -1, 0, 1, -1, 1, -1},
    dz[10] = {0, 0, 0, 0, 1, -1};
long long x, y, mod = 1000000007, mod2 = 1000000009, mod3 = 2017;
double pi = 3.14159265;
P u[1];
queue<int> q;
bool as(P a, P b) {
  if (a.x != b.x) return a.x < b.x;
  return a.y < b.y;
}
int main() {
  scanf("%d %d %d", &a, &b, &c);
  n = a;
  m = b;
  x = (long long)a * b;
  if (x % c) {
    if (c % 2 == 0 && x % c == c / 2) {
      for (long long i = 2; i * i <= c; i++) {
        for (; c % i == 0 && (a % i == 0 || b % i == 0); c /= i) {
          if (a % i == 0)
            a /= i;
          else
            b /= i;
        }
      }
      if (c > 2) {
        if (a % (c / 2) == 0)
          a /= (c / 2);
        else if (b % (c / 2) == 0)
          b /= (c / 2);
      }
      puts("YES");
      printf("0 0\n%d 0\n0 %d", a, b);
      return 0;
    } else {
      puts("NO");
      return 0;
    }
  }
  for (long long i = 2; i * i <= c; i++) {
    for (; c % i == 0 && (a % i == 0 || b % i == 0); c /= i) {
      if (a % i == 0)
        a /= i;
      else
        b /= i;
    }
  }
  if (c > 1) {
    if (a % c == 0)
      a /= c;
    else if (b % c == 0)
      b /= c;
  }
  if (n > a)
    a *= 2;
  else
    b *= 2;
  puts("YES");
  printf("0 0\n%d 0\n0 %d", a, b);
}
