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, N, M, K;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &k);
  N = n;
  M = m;
  K = k;
  if (k == 1) {
    puts("NO");
    return 0;
  }
  long long d = gcd(n * m, k);
  long long t = k / d, w = n * m / d;
  if (!(n % d))
    n /= d;
  else if (!(m % d))
    m /= d;
  if (t == 1) {
    if (n * m / 2 > w) {
      long long w1 = gcd(n, k);
      n /= w1;
      if (n * m / 2 > w) {
        long long w2 = gcd(m, k);
        m /= w2;
      }
    }
    if (n * m / 2 > w) {
      if (!(n & 1)) n /= 2;
      if (n * m / 2 > w) {
        if (!(m & 1)) m /= 2;
      }
    }
    if (n * 2 <= N)
      n *= 2;
    else if (m * 2 <= M)
      m *= 2;
    if (w <= N && M >= 2) {
      puts("YES");
      printf("0 0\n%I64d 0\n0 2\n", w);
    } else if (w <= M && N >= 2) {
      puts("YES");
      printf("0 0\n0 %I64d\n2 0\n", w);
    } else if (w > N && w > M) {
      puts("YES");
      printf("0 0\n%I64d 0\n%I64d %I64d\n", n, n, m);
    } else
      puts("NO");
  } else if (t == 2) {
    puts("YES");
    if (n * m > w) {
      long long w1 = gcd(n, k);
      n /= w1;
      if (n * m > w) {
        long long w2 = gcd(m, k);
        m /= w2;
      }
    }
    if (n * m > w) {
      if (!(n & 1)) n /= 2;
      if (n * m > w) {
        if (!(m & 1)) m /= 2;
      }
    }
    if (w <= N) {
      printf("0 0\n%I64d 0\n0 1\n", w);
    } else if (w <= M) {
      printf("0 0\n0 %I64d\n1 0\n", w);
    } else
      printf("0 0\n%I64d 0\n%I64d %I64d\n", n, n, m);
  } else
    puts("NO");
  return 0;
}
