#include <bits/stdc++.h>
using namespace std;
vector<long long int> ext_gcd(long long int a, long long int b) {
  if (b == 0) {
    return {a, 1, 0};
  }
  vector<long long int> res = ext_gcd(b, a % b);
  return {res[0], res[2], res[1] - (a / b) * res[2]};
}
signed main() {
  long long int a, b, c;
  cin >> a >> b >> c;
  if (a == 0 && b == 0) {
    if (c == 0)
      cout << 0 << " " << 0 << "\n";
    else
      cout << -1 << "\n";
    return 0;
  } else if (a == 0) {
    if (c == 0)
      cout << 0 << " " << 0 << "\n";
    else {
      if (abs(c) % abs(b) == 0)
        cout << 0 << " " << -c / b << "\n";
      else
        cout << -1 << "\n";
    }
    return 0;
  } else if (b == 0) {
    if (c == 0)
      cout << 0 << " " << 0 << "\n";
    else {
      if (abs(c) % abs(a) == 0)
        cout << -c / a << " " << 0 << "\n";
      else
        cout << -1 << "\n";
    }
    return 0;
  }
  c = -c;
  vector<long long int> ans = ext_gcd(a, b);
  if (c % ans[0] != 0)
    cout << -1;
  else {
    long long int kol = c / ans[0];
    ans[1] *= kol;
    ans[2] *= kol;
    long long int k = (-ans[1] * ans[0]) / b;
    for (long long int i = k - 1000; i < k + 1000; i++) {
      long long int nx = ans[1] + i * (b / ans[0]);
      long long int ny = ans[2] - i * (a / ans[0]);
      if (nx >= -5e18 && nx <= 5e18 && ny >= -5e18 && ny < 5e18) {
        ans[1] = ans[1] + i * (b / ans[0]);
        ans[2] = ans[2] - i * (a / ans[0]);
        break;
      }
    }
    if (ans[1] >= -5e18 && ans[1] <= 5e18 && ans[2] >= -5e18 && ans[2] < 5e18)
      cout << ans[1] << " " << ans[2] << "\n";
    else
      cout << -1 << "\n";
  }
  return 0;
}

### 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;
vector<long long int> ext_gcd(long long int a, long long int b) {
  if (b == 0) {
    return {a, 1, 0};
  }
  vector<long long int> res = ext_gcd(b, a % b);
  return {res[0], res[2], res[1] - (a / b) * res[2]};
}
signed main() {
  long long int a, b, c;
  cin >> a >> b >> c;
  if (a == 0 && b == 0) {
    if (c == 0)
      cout << 0 << " " << 0 << "\n";
    else
      cout << -1 << "\n";
    return 0;
  } else if (a == 0) {
    if (c == 0)
      cout << 0 << " " << 0 << "\n";
    else {
      if (abs(c) % abs(b) == 0)
        cout << 0 << " " << -c / b << "\n";
      else
        cout << -1 << "\n";
    }
    return 0;
  } else if (b == 0) {
    if (c == 0)
      cout << 0 << " " << 0 << "\n";
    else {
      if (abs(c) % abs(a) == 0)
        cout << -c / a << " " << 0 << "\n";
      else
        cout << -1 << "\n";
    }
    return 0;
  }
  c = -c;
  vector<long long int> ans = ext_gcd(a, b);
  if (c % ans[0] != 0)
    cout << -1;
  else {
    long long int kol = c / ans[0];
    ans[1] *= kol;
    ans[2] *= kol;
    long long int k = (-ans[1] * ans[0]) / b;
    for (long long int i = k - 1000; i < k + 1000; i++) {
      long long int nx = ans[1] + i * (b / ans[0]);
      long long int ny = ans[2] - i * (a / ans[0]);
      if (nx >= -5e18 && nx <= 5e18 && ny >= -5e18 && ny < 5e18) {
        ans[1] = ans[1] + i * (b / ans[0]);
        ans[2] = ans[2] - i * (a / ans[0]);
        break;
      }
    }
    if (ans[1] >= -5e18 && ans[1] <= 5e18 && ans[2] >= -5e18 && ans[2] < 5e18)
      cout << ans[1] << " " << ans[2] << "\n";
    else
      cout << -1 << "\n";
  }
  return 0;
}

```