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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int gcd_util(long long int a, long long int b) {
  return b ? gcd_util(b, a % b) : a;
}
long long int gcd(long long int a, long long int b) {
  return gcd_util(abs(a), abs(b));
}
pair<long long int, long long int> extended_gcd_util(long long int a,
                                                     long long int b) {
  if (b == 0) {
    return {1, 0};
  }
  pair<long long int, long long int> p = extended_gcd_util(b, a % b);
  long long int x = p.first, y = p.second;
  return {y, x - (a / b) * y};
}
pair<long long int, long long int> extended_gcd(long long int a,
                                                long long int b) {
  pair<long long int, long long int> p = extended_gcd_util(abs(a), abs(b));
  if (a < 0) {
    p.first *= (-1);
  }
  if (b < 0) {
    p.second *= (-1);
  }
  return p;
}
void linearDiophantine_OneSol(long long int a, long long int b,
                              long long int c) {
  long long int g = gcd(a, b);
  if ((c % g) != 0) {
    cout << "-1\n";
    return;
  }
  long long int x, y;
  pair<long long int, long long int> p = extended_gcd(a, b);
  x = p.first;
  y = p.second;
  x *= (c / g);
  y *= (c / g);
  cout << x << " " << y << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int a, b, c;
  cin >> a >> b >> c;
  c *= (-1);
  linearDiophantine_OneSol(a, b, c);
  return 0;
}