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 get_num() {
  long long num = 0;
  char c;
  bool flag = false;
  while ((c = getchar()) == ' ' || c == '\r' || c == '\n')
    ;
  if (c == '-')
    flag = true;
  else
    num = c - '0';
  while (isdigit(c = getchar())) num = num * 10 + c - '0';
  return (flag ? -1 : 1) * num;
}
long long n, m, k;
long long sum;
const int mx = 1e5 + 5;
bool vis[mx];
long long pr[mx];
bool isp = true;
int cnt = 0;
void euler() {
  for (int i = 2; i <= 1e5; ++i) {
    if (!vis[i]) pr[++cnt] = i;
    for (int j = 1; pr[j] * i <= 100000 && j <= cnt; ++j) {
      vis[pr[j] * i] = false;
      if (i % pr[j] == 0) break;
    }
  }
  return;
}
bool pn = false, pm = false;
int main() {
  memset(vis, 0, sizeof(vis));
  memset(pr, 0, sizeof(pr));
  cnt = 0;
  n = get_num();
  m = get_num();
  k = get_num();
  sum = n * m;
  euler();
  if (sum % k == 0) {
    printf("YES\n");
    printf("0 0\n");
    sum /= k;
    for (int i = 2; i <= sqrt(k); ++i) {
      if (k % i == 0) {
        isp = false;
        break;
      }
    }
    if (isp) {
      if (n % k == 0) {
        n /= k;
        pn = true;
      } else if (m % k == 0) {
        m /= k;
        pm = true;
      }
      k /= k;
    } else {
      for (int i = 1; i <= cnt; ++i) {
        while (k % pr[i] == 0) {
          k /= pr[i];
          if (n % pr[i] == 0) {
            n /= pr[i];
            pn = true;
          } else {
            m /= pr[i];
            pm = true;
          }
        }
      }
    }
    if (n % k == 0 && k != 1) {
      n /= k;
      pn = true;
    } else if (m % k == 0 && k != 1) {
      m /= k;
      pm = true;
    }
    if (pn)
      n *= 2;
    else
      m *= 2;
    printf("%I64d 0\n", n);
    printf("0 %I64d\n", m);
  } else if (k % 2 == 0 && sum % (k / 2) == 0) {
    printf("YES\n");
    printf("0 0\n");
    k /= 2;
    for (int i = 2; i <= sqrt(k); ++i) {
      if (k % i == 0) {
        isp = false;
        break;
      }
    }
    if (isp) {
      if (n % k == 0) {
        n /= k;
        pn = true;
      } else if (m % k == 0) {
        m /= k;
        pm = true;
      }
      k /= k;
    } else {
      for (int i = 1; i <= cnt; ++i) {
        while (k % pr[i] == 0) {
          k /= pr[i];
          if (n % pr[i] == 0)
            n /= pr[i];
          else
            m /= pr[i];
        }
      }
    }
    if (n % k == 0 && k != 1) {
      n /= k;
      pn = true;
    } else if (m % k == 0 && k != 1) {
      m /= k;
      pm = true;
    }
    printf("%I64d 0\n", n);
    printf("0 %I64d\n", m);
  } else
    printf("NO\n");
  return 0;
}
