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;
inline int Gcd(int X, int Y) { return Y ? Gcd(Y, X % Y) : X; }
inline long long Gcd(long long X, long long Y) { return Y ? Gcd(Y, X % Y) : X; }
inline int Pow(int B, long long E, int M) {
  if (!(B %= M)) return 0;
  int A = 1;
  for (; E; E >>= 1, B = (long long)B * B % M)
    E& 1 ? A = (long long)A * B % M : 0;
  return A;
}
inline long long Pow(long long base, long long exp, long long _mod) {
  if (!(base %= _mod)) return 0;
  long long _ans = 1;
  for (; exp; exp >>= 1, base = base * base % _mod)
    exp& 1 ? _ans = _ans * base % _mod : 0;
  return _ans;
}
const int INF = 0x3f3f3f3f;
int n, m, k;
int a[300001], b[300001];
int h[300001], nxt[600001], to[600001], tot;
inline void ins(int x, int y) {
  nxt[++tot] = h[x];
  to[tot] = y;
  h[x] = tot;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  if ((long long)n * m * 2 % k != 0) return puts("NO"), 0;
  long long S = (long long)n * m * 2 / k;
  puts("YES");
  if (k % 2 == 0) {
    long long A1 = Gcd((long long)n, S);
    long long A2 = S / A1;
    printf("0 0\n%lld 0\n0 %lld\n", A1, A2);
    return 0;
  } else {
    long long A1 = Gcd((long long)n, S);
    long long A2 = S / A1;
    if (A1 * 2 <= n) A1 *= 2, A2 /= 2;
    printf("0 0\n%lld 0\n0 %lld\n", A1, A2);
  }
  return 0;
}
