#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long lnf = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const double eps = 1e-4;
const int N = 1e6 + 10;
const int M = 1e6 + 10;
const int mod = 19940417;
long long a, b, c;
long long exgcd(long long a, long long b, long long& x, long long& y) {
  long long res = a;
  if (!b)
    x = 1, y = 0;
  else {
    res = exgcd(b, a % b, x, y);
    long long temp = x;
    x = y;
    y = temp - a / b * y;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> a >> b >> c;
  long long x, y;
  long long g = exgcd(a, b, x, y);
  if (c % g == 0)
    cout << -x * c / g << " " << -y * c / g << '\n';
  else
    cout << -1 << '\n';
  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;
const int inf = 0x3f3f3f3f;
const long long lnf = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const double eps = 1e-4;
const int N = 1e6 + 10;
const int M = 1e6 + 10;
const int mod = 19940417;
long long a, b, c;
long long exgcd(long long a, long long b, long long& x, long long& y) {
  long long res = a;
  if (!b)
    x = 1, y = 0;
  else {
    res = exgcd(b, a % b, x, y);
    long long temp = x;
    x = y;
    y = temp - a / b * y;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> a >> b >> c;
  long long x, y;
  long long g = exgcd(a, b, x, y);
  if (c % g == 0)
    cout << -x * c / g << " " << -y * c / g << '\n';
  else
    cout << -1 << '\n';
  return 0;
}

```