#include <bits/stdc++.h>
using namespace std;
long long a, b, c, g, x, y;
inline long long fixMod(long long a, long long m) { return (a % m + m) % m; }
inline long long floor(long long a, long long b) {
  return (a - fixMod(a, b)) / b;
}
void getNewTerm(long long& g0, long long& g1, long long q) {
  long long t = g0 - q * g1;
  g0 = g1;
  g1 = t;
}
long long eGCD(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long g = eGCD(b, a % b, y, x);
  y -= (a / b) * x;
  return g;
}
bool solveLDE(long long a, long long b, long long c, long long& x, long long& y,
              long long& g) {
  g = eGCD(a, b, x, y);
  x *= c / g;
  y *= c / g;
  return !(c % g);
}
int main() {
  cin >> a >> b >> c;
  bool sol = solveLDE(a, b, -1 * c, x, y, g);
  if (!sol) {
    cout << "-1";
    return 0;
  } else {
    cout << x << " " << y;
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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 a, b, c, g, x, y;
inline long long fixMod(long long a, long long m) { return (a % m + m) % m; }
inline long long floor(long long a, long long b) {
  return (a - fixMod(a, b)) / b;
}
void getNewTerm(long long& g0, long long& g1, long long q) {
  long long t = g0 - q * g1;
  g0 = g1;
  g1 = t;
}
long long eGCD(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long g = eGCD(b, a % b, y, x);
  y -= (a / b) * x;
  return g;
}
bool solveLDE(long long a, long long b, long long c, long long& x, long long& y,
              long long& g) {
  g = eGCD(a, b, x, y);
  x *= c / g;
  y *= c / g;
  return !(c % g);
}
int main() {
  cin >> a >> b >> c;
  bool sol = solveLDE(a, b, -1 * c, x, y, g);
  if (!sol) {
    cout << "-1";
    return 0;
  } else {
    cout << x << " " << y;
  }
  return 0;
}

```