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;
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  if (2l * m * n % k != 0) {
    printf("NO\n");
    return 0;
  }
  long long inN = n, inM = m, inK = k;
  map<long long, int> nF, mF, kF;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) nF[i]++, n /= i;
  }
  nF[n]++;
  for (long long i = 2; i * i <= m; ++i) {
    while (m % i == 0) mF[i]++, m /= i;
  }
  mF[m]++;
  for (long long i = 2; i * i <= k; ++i) {
    while (k % i == 0) kF[i]++, k /= i;
  }
  kF[k]++;
  bool is2Used = false;
  long long nAns = inN;
  long long mAns = inM;
  for (auto it = kF.begin(); it != kF.end(); it++) {
    ;
    if (false == is2Used && it->first == 2) {
      ;
      is2Used = true;
      it->second--;
    }
    if (nF[it->first] != 0) {
      ;
      int reduce = min(nF[it->first], it->second);
      it->second -= reduce;
      while (reduce--) {
        nAns /= it->first;
      };
    }
  }
  if (false == is2Used && 2 * nAns <= inN) {
    nAns *= 2;
    is2Used = true;
  }
  for (auto it = kF.begin(); it != kF.end(); it++) {
    ;
    if (false == is2Used && it->first == 2) {
      ;
      is2Used = true;
      it->second--;
    }
    if (mF[it->first] != 0) {
      int reduce = min(mF[it->first], it->second);
      it->second -= reduce;
      while (reduce--) {
        mAns /= it->first;
      }
    };
  }
  if (false == is2Used && 2 * mAns <= inM) {
    mAns *= 2;
    is2Used = true;
  }
  printf("YES\n");
  printf("0 0\n");
  printf("%lld 0\n", nAns);
  printf("0 %lld\n", mAns);
  return 0;
}
