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 mod = 1000000007;
long long mod2 = 998244353;
long long mod3 = 1000003;
long long mod4 = 998244853;
long long mod5 = 1000000009;
long long inf = 1LL << 60;
double pi = 3.141592653589793238462643383279L;
double eps = 1e-14;
int dh[4] = {1, -1, 0, 0};
int dw[4] = {0, 0, 1, -1};
int ddh[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddw[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long c = gcd(a, b);
  return a * b / c;
}
long long Pow(long long n, long long k) {
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k & 1) ret *= now;
    now *= now;
    k /= 2;
  }
  return ret;
}
long long beki(long long n, long long k, long long md) {
  long long ret = 1;
  long long now = n;
  now %= md;
  while (k > 0) {
    if (k % 2 == 1) {
      ret *= now;
      ret %= md;
    }
    now *= now;
    now %= md;
    k /= 2;
  }
  return ret;
}
long long gyaku(long long n, long long md) { return beki(n, md - 2, md); }
long long popcount(long long n) {
  long long ret = 0;
  long long u = n;
  while (u > 0) {
    ret += u % 2;
    u /= 2;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> v, w;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      v.push_back(i);
      if (i * i < n) v.push_back(n / i);
    }
  }
  for (long long i = 1; i * i <= m; i++) {
    if (m % i == 0) {
      w.push_back(i);
      if (i * i < m) w.push_back(m / i);
    }
  }
  if (k % 2 == 0) {
    k /= 2;
    long long r = 0;
    for (long long i = 0; i < v.size(); i++) {
      for (long long j = 0; j < w.size(); j++) {
        long long f = v[i], g = w[j];
        if (f * g == k) {
          n /= f;
          m /= g;
          r = 1;
          break;
        }
      }
      if (r) break;
    }
    if (r == 0) {
      cout << "NO" << endl;
      return 0;
    }
    cout << "YES" << endl;
    cout << 0 << " " << 0 << endl
         << n << " " << 0 << endl
         << 0 << " " << m << endl;
  } else {
    long long r = 0;
    for (long long i = 0; i < v.size(); i++) {
      for (long long j = 0; j < w.size(); j++) {
        long long f = v[i], g = w[j];
        if (f * g == k) {
          n /= f;
          m /= g;
          r = 1;
          if (f == 1)
            m *= 2;
          else
            n *= 2;
          break;
        }
      }
      if (r) break;
    }
    if (r == 0) {
      cout << "NO" << endl;
      return 0;
    }
    cout << "YES" << endl;
    cout << 0 << " " << 0 << endl
         << n << " " << 0 << endl
         << 0 << " " << m << endl;
  }
}
