#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int mod = 998244353;
const int maxn = 2e6 + 7;
const int lim = 800667;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
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 d = exgcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - (a / b) * y;
  return d;
}
long long x, y;
int main() {
  long long a, b, c;
  scanf("%lld%lld%lld", &a, &b, &c);
  if (c % gcd(a, b) != 0)
    printf("-1\n");
  else {
    long long G = exgcd(a, b, x, y);
    long long res1 = -x * c / G;
    long long res2 = -y * c / G;
    printf("%lld %lld\n", res1, res2);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int mod = 998244353;
const int maxn = 2e6 + 7;
const int lim = 800667;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
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 d = exgcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - (a / b) * y;
  return d;
}
long long x, y;
int main() {
  long long a, b, c;
  scanf("%lld%lld%lld", &a, &b, &c);
  if (c % gcd(a, b) != 0)
    printf("-1\n");
  else {
    long long G = exgcd(a, b, x, y);
    long long res1 = -x * c / G;
    long long res2 = -y * c / G;
    printf("%lld %lld\n", res1, res2);
  }
  return 0;
}

```