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>
#pragma warning(disable : 4996)
using namespace std;
const int INF = 0x3f3f3f3f;
const long long int INFLL = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e6 + 3;
struct Exgcd {
  long long exgcd(long long a, long long b, long long& x, long long& y) {
    if (!b) {
      x = 1;
      y = 0;
      return a;
    }
    long long int t = exgcd(b, a % b, y, x);
    y -= x * (a / b);
    return t;
  }
  pair<long long, long long> get(long long a, long long b, long long c) {
    long long x = 0;
    long long y = 0;
    long long t = exgcd(a, b, x, y);
    if (c % t)
      return make_pair((long long int)5e18 + 1, (long long int)5e18 + 1);
    x *= c / t;
    y *= c / t;
    return make_pair(x, y);
  }
  long long get(long long a, long long mod) {
    long long x = 0;
    long long y = 0;
    exgcd(a, mod, x, y);
    return (x % mod + mod) % mod;
  }
};
long long fpow(long long a, long long n, long long mod) {
  if (n == 0)
    return 1;
  else if (n & 1)
    return a * fpow(a, n - 1, mod) % mod;
  else {
    long long num = fpow(a, n / 2, mod) % mod;
    return num * num % mod;
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int a, b, c;
  Exgcd e;
  cin >> a >> b >> c;
  pair<long long int, long long int> ans = e.get(a, b, -c);
  if (abs(ans.first) > (long long int)5e18 ||
      abs(ans.second) > (long long int)5e18)
    cout << -1;
  else
    cout << ans.first << ' ' << ans.second;
}