#include <bits/stdc++.h>
using namespace std;
long long extended_gcd(long long a, long long b, long long& x, long long& y) {
  x = 1;
  y = 0;
  long long xnew = 0;
  long long ynew = 1;
  while (b) {
    long long r = a / b;
    long long t = a - r * b;
    a = b;
    b = t;
    t = x - r * xnew;
    x = xnew;
    xnew = t;
    t = y - r * ynew;
    y = ynew;
    ynew = t;
  }
  return a;
}
bool solve_diphontine_equation(long long a, long long b, long long c,
                               long long& g, long long& x, long long& y) {
  g = extended_gcd(a, b, x, y);
  if (c % g) return false;
  long long m = c / g;
  x *= m;
  y *= m;
  return true;
}
int main() {
  long long a, b, c, g, x, y;
  cin >> a >> b >> c;
  if (!solve_diphontine_equation(a, b, -c, g, x, y)) {
    cout << "-1\n";
  } else
    cout << x << " " << y << endl;
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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;
long long extended_gcd(long long a, long long b, long long& x, long long& y) {
  x = 1;
  y = 0;
  long long xnew = 0;
  long long ynew = 1;
  while (b) {
    long long r = a / b;
    long long t = a - r * b;
    a = b;
    b = t;
    t = x - r * xnew;
    x = xnew;
    xnew = t;
    t = y - r * ynew;
    y = ynew;
    ynew = t;
  }
  return a;
}
bool solve_diphontine_equation(long long a, long long b, long long c,
                               long long& g, long long& x, long long& y) {
  g = extended_gcd(a, b, x, y);
  if (c % g) return false;
  long long m = c / g;
  x *= m;
  y *= m;
  return true;
}
int main() {
  long long a, b, c, g, x, y;
  cin >> a >> b >> c;
  if (!solve_diphontine_equation(a, b, -c, g, x, y)) {
    cout << "-1\n";
  } else
    cout << x << " " << y << endl;
  return 0;
}

```