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;
const int maxn = 1e5 + 10;
vector<long long> A, B;
map<long long, int> ma, mb;
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  if (n * m * 2 % k != 0) {
    puts("NO");
    return 0;
  } else if (!(k & 1)) {
    long long nn = n, mm = m;
    long long x = n;
    for (long long i = 2; i * i <= x; ++i) {
      if (x % i == 0) {
        A.push_back(i);
        while (x % i == 0) x /= i, ma[i]++;
      }
    }
    if (x > 1) A.push_back(x), ma[x]++;
    x = m * 2;
    m *= 2;
    for (long long i = 2; i * i <= x; ++i) {
      if (x % i == 0) {
        B.push_back(i);
        while (x % i == 0) x /= i, mb[i]++;
      }
    }
    if (x > 1) B.push_back(x), mb[x]++;
    x = k;
    for (long long i = 2; i * i <= x; ++i) {
      if (x % i == 0) {
        while (x % i == 0) {
          if (mb[i] > 0)
            mb[i]--, m /= i;
          else
            ma[i]--, n /= i;
          x /= i;
        }
      }
    }
    if (x > 1) {
      if (mb[x] > 0)
        mb[x]--, m /= x;
      else
        ma[x]--, n /= x;
    }
    if (n <= nn && m <= mm) {
      puts("YES");
      puts("0 0");
      printf("0 %lld\n", m);
      printf("%lld 0\n", n);
    } else
      puts("NO");
  } else {
    long long nn = n, mm = m;
    long long x = n;
    for (long long i = 2; i * i <= x; ++i) {
      if (x % i == 0) {
        A.push_back(i);
        while (x % i == 0) x /= i, ma[i]++;
      }
    }
    if (x > 1) A.push_back(x), ma[x]++;
    x = m;
    for (long long i = 2; i * i <= x; ++i) {
      if (x % i == 0) {
        B.push_back(i);
        while (x % i == 0) x /= i, mb[i]++;
      }
    }
    if (x > 1) B.push_back(x), mb[x]++;
    x = k;
    for (long long i = 2; i * i <= x; ++i) {
      if (x % i == 0) {
        while (x % i == 0) {
          if (mb[i] > 0)
            mb[i]--, m /= i;
          else
            ma[i]--, n /= i;
          x /= i;
        }
      }
    }
    if (x > 1) {
      if (mb[x] > 0)
        mb[x]--, m /= x;
      else
        ma[x]--, n /= x;
    }
    if (n * 2 <= nn)
      n *= 2;
    else
      m *= 2;
    if (n <= nn && m <= mm) {
      puts("YES");
      puts("0 0");
      printf("0 %lld\n", m);
      printf("%lld 0\n", n);
    } else
      puts("NO");
  }
  return 0;
}
