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;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ll = long long int;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vd = vector<double>;
using vvd = vector<vd>;
using vvvd = vector<vvd>;
using P = pair<int, int>;
using Pll = pair<ll, ll>;
using cdouble = complex<double>;
const double eps = 1e-9;
const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
bool nearlyeq(double x, double y) { return abs(x - y) < eps; }
bool inrange(ll x, ll t) { return x >= 0 && x < t; }
bool inrange(vll xs, ll t) {
  for (auto x : xs)
    if (!(x >= 0 && x < t)) return false;
  return true;
}
int ceillog2(ll x) {
  int ret = 0;
  x--;
  while (x > 0) {
    ret++;
    x >>= 1;
  }
  return ret;
}
ll rndf(double x) { return (ll)(x + (x >= 0 ? 0.5 : -0.5)); }
ll floorsqrt(ll x) {
  ll m = (ll)sqrt((double)x);
  return m + (m * m <= x ? 0 : -1);
}
ll ceilsqrt(ll x) {
  ll m = (ll)sqrt((double)x);
  return m + (x <= m * m ? 0 : 1);
}
ll rnddiv(ll a, ll b) { return (a / b + (a % b * 2 >= b ? 1 : 0)); }
ll ceildiv(ll a, ll b) { return (a / b + (a % b == 0 ? 0 : 1)); }
ll gcd(ll m, ll n) {
  if (n == 0)
    return m;
  else
    return gcd(n, m % n);
}
ll lcm(ll m, ll n) { return m * n / gcd(m, n); }
void solve(ll nn, ll mm, ll k, bool rev_flag = false) {
  ll n = nn;
  ll m = mm;
  if (k % 2 == 0)
    k /= 2;
  else
    n *= 2;
  ll gn = gcd(n, k);
  n /= gn;
  k /= gn;
  ll gm = gcd(m, k);
  m /= gm;
  k /= gm;
  if (k == 1 && n <= nn) {
    if (rev_flag) {
      cout << "YES" << endl;
      cout << 0 << " " << 0 << endl;
      cout << 0 << " " << n << endl;
      cout << m << " " << 0 << endl;
    } else {
      cout << "YES" << endl;
      cout << 0 << " " << 0 << endl;
      cout << n << " " << 0 << endl;
      cout << 0 << " " << m << endl;
    }
    exit(0);
  }
}
int main() {
  ll n, m, k;
  cin >> n >> m >> k;
  solve(n, m, k);
  solve(m, n, k, true);
  cout << "NO" << endl;
}
