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 int gcd(long long int a, long long int b) {
  while (b) {
    long long int t = a % b;
    a = b;
    b = t;
  }
  return a;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n, m, k;
  cin >> n >> m >> k;
  if (k == 1) {
    cout << "NO";
    return 0;
  }
  long long int num = 1ll * n * m * 2ll;
  long long int den = k;
  long long int g = gcd(num, den);
  num /= g;
  den /= g;
  if (den != 1) {
    cout << "NO";
    return 0;
  }
  if (num > 1ll * n * m) {
    cout << "NO";
    return 0;
  }
  if (num == 1ll * n * m) {
    long long int ans_a = n, ans_b = m;
    cout << "YES\n";
    cout << 0 << ' ' << 0 << '\n';
    cout << ans_a << ' ' << 0 << '\n';
    cout << 0 << ' ' << ans_b << '\n';
    return 0;
  }
  bool f = 0;
  if (g % 2 == 0) {
    f = 1;
    g >>= 1;
  }
  long long int a = gcd(n, g);
  long long int b = g / a;
  long long int ans_a = n / a, ans_b = m / gcd(b, m);
  if (!f && a >= 2) {
    ans_a *= 2ll;
  } else if (!f && b >= 2) {
    ans_b *= 2ll;
  } else if (!f) {
    cout << "NO";
    return 0;
  }
  if (ans_a * ans_b != num) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  cout << 0 << ' ' << 0 << '\n';
  cout << ans_a << ' ' << 0 << '\n';
  cout << 0 << ' ' << ans_b << '\n';
  return 0;
}
