#include <bits/stdc++.h>
using namespace std;
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0)
    if (b % 2 == 0) {
      a *= a;
      a %= m;
      b /= 2;
    } else {
      ans *= a;
      ans %= m;
      b--;
    }
  return ans;
}
int fi(int n) {
  long long ans = n;
  set<int> s;
  s.clear();
  for (long long p = 2; p * p <= n; p++) {
    if (n % p == 0) s.insert(p);
    while (n % p == 0) n /= p;
  };
  if (n > 1) s.insert(n);
  for (set<int>::iterator it = s.begin(); it != s.end(); it++) {
    int p = *it;
    ans *= p - 1;
    ans /= p;
  }
  return ans;
}
long long rev(long long a, long long m) { return pow(a, fi(m) - 1, m); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main(void) {
  long long A, B, C, D;
  cin >> A >> B >> C;
  D = gcd(A, B);
  A /= D;
  B /= D;
  if (C % D != 0) {
    cout << "-1" << endl;
    return 0;
  }
  C /= D;
  if (A == 0) {
    if (C % B == 0)
      cout << "0 " << -C / B << endl;
    else
      cout << "-1" << endl;
    return 0;
  }
  if (B == 0) {
    if (C % A == 0)
      cout << -C / A << " 0" << endl;
    else
      cout << "-1" << endl;
    return 0;
  }
  if (B < 0) {
    A *= -1;
    B *= -1;
    C *= -1;
  }
  long long x = ((rev((A % B + B) % B, B) * -C) % B + B) % B;
  long long y = (-x * A - C) / B;
  cout << x << ' ' << y << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0)
    if (b % 2 == 0) {
      a *= a;
      a %= m;
      b /= 2;
    } else {
      ans *= a;
      ans %= m;
      b--;
    }
  return ans;
}
int fi(int n) {
  long long ans = n;
  set<int> s;
  s.clear();
  for (long long p = 2; p * p <= n; p++) {
    if (n % p == 0) s.insert(p);
    while (n % p == 0) n /= p;
  };
  if (n > 1) s.insert(n);
  for (set<int>::iterator it = s.begin(); it != s.end(); it++) {
    int p = *it;
    ans *= p - 1;
    ans /= p;
  }
  return ans;
}
long long rev(long long a, long long m) { return pow(a, fi(m) - 1, m); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main(void) {
  long long A, B, C, D;
  cin >> A >> B >> C;
  D = gcd(A, B);
  A /= D;
  B /= D;
  if (C % D != 0) {
    cout << "-1" << endl;
    return 0;
  }
  C /= D;
  if (A == 0) {
    if (C % B == 0)
      cout << "0 " << -C / B << endl;
    else
      cout << "-1" << endl;
    return 0;
  }
  if (B == 0) {
    if (C % A == 0)
      cout << -C / A << " 0" << endl;
    else
      cout << "-1" << endl;
    return 0;
  }
  if (B < 0) {
    A *= -1;
    B *= -1;
    C *= -1;
  }
  long long x = ((rev((A % B + B) % B, B) * -C) % B + B) % B;
  long long y = (-x * A - C) / B;
  cout << x << ' ' << y << endl;
  return 0;
}

```