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;
map<long long, long long> d;
vector<pair<long long, long long> > e;
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  if ((n * m * 2) % k != 0) {
    cout << "NO";
    return 0;
  }
  if (k == 2) {
    int x = n;
    int y = m;
    cout << "YES" << endl;
    cout << 0 << " " << 0 << endl;
    cout << 0 << " " << y << endl;
    cout << x << " " << 0 << endl;
    return 0;
  }
  d[2]++;
  long long n1 = n;
  for (long long i = 2; i <= (long long)sqrt(n1); i++) {
    while (n % i == 0) {
      n /= i;
      d[i]++;
    }
  }
  d[n]++;
  n = n1;
  long long m1 = m;
  for (long long i = 2; i <= (long long)sqrt(m1); i++) {
    while (m % i == 0) {
      m /= i;
      d[i]++;
    }
  }
  d[m]++;
  m = m1;
  long long k1 = k;
  for (long long i = 2; i <= (long long)sqrt(k1); i++) {
    while (k % i == 0) {
      k /= i;
      d[i]--;
    }
  }
  d[k]--;
  k = k1;
  long long x = 1, y = 1;
  e.clear();
  for (auto it = d.begin(); it != d.end(); it++) {
    long long y1 = it->first, y2 = it->second;
    if (y1 < 2) continue;
    e.push_back({y1, y2});
  }
  reverse(e.begin(), e.end());
  int num = 0;
  for (long long i = 0; i < (long long)e.size(); i++) {
    for (long long j = 0; j < e[i].second; j++) {
      num++;
      if (y * e[i].first > m) {
        x *= e[i].first;
        continue;
      }
      if (x * e[i].first > n) {
        y *= e[i].first;
        continue;
      }
      if (num % 2 == 1)
        x *= e[i].first;
      else
        y *= e[i].first;
    }
  }
  if (x > n || y > m) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  cout << 0 << " " << 0 << endl;
  cout << 0 << " " << y << endl;
  cout << x << " " << 0 << endl;
}
