#include <bits/stdc++.h>
using namespace std;
pair<long long, pair<long long, long long> > extendedEuclid(long long a,
                                                            long long b) {
  if (a == 0) return make_pair(b, make_pair(0, 1));
  pair<long long, pair<long long, long long> > p;
  p = extendedEuclid(b % a, a);
  return make_pair(
      p.first,
      make_pair(p.second.second - p.second.first * (b / a), p.second.first));
}
int main() {
  long long a, b, c;
  scanf("%lld%lld%lld", &a, &b, &c);
  long long x = extendedEuclid(a, b).second.first,
            y = extendedEuclid(a, b).second.second;
  c = -c;
  long long s = a * x + b * y;
  if (c % s == 0)
    printf("%lld %lld", c / s * x, c / s * y);
  else
    printf("%d", -1);
}

### Prompt

Develop a solution in cpp to the problem described below:

A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from  - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.

Input

The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.

Output

If the required point exists, output its coordinates, otherwise output -1.

Examples

Input

2 5 3


Output

6 -3

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
pair<long long, pair<long long, long long> > extendedEuclid(long long a,
                                                            long long b) {
  if (a == 0) return make_pair(b, make_pair(0, 1));
  pair<long long, pair<long long, long long> > p;
  p = extendedEuclid(b % a, a);
  return make_pair(
      p.first,
      make_pair(p.second.second - p.second.first * (b / a), p.second.first));
}
int main() {
  long long a, b, c;
  scanf("%lld%lld%lld", &a, &b, &c);
  long long x = extendedEuclid(a, b).second.first,
            y = extendedEuclid(a, b).second.second;
  c = -c;
  long long s = a * x + b * y;
  if (c % s == 0)
    printf("%lld %lld", c / s * x, c / s * y);
  else
    printf("%d", -1);
}

```