#include <bits/stdc++.h>
int main() {
  long long x[2] = {0, -1};
  long long y[2] = {-1, 0};
  int a, b, c, q, r, p = 0;
  scanf("%d %d %d", &a, &b, &c);
  while (b) {
    p = !p;
    r = a % b;
    q = (a - r) / b;
    x[p] -= q * x[!p];
    y[p] -= q * y[!p];
    a = b;
    b = r;
  }
  if (c % a)
    printf("-1\n");
  else
    printf("%lli %lli\n", x[!p] * c / a, y[!p] * c / a);
  return 0;
}

### Prompt

Please create a solution in CPP to 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>
int main() {
  long long x[2] = {0, -1};
  long long y[2] = {-1, 0};
  int a, b, c, q, r, p = 0;
  scanf("%d %d %d", &a, &b, &c);
  while (b) {
    p = !p;
    r = a % b;
    q = (a - r) / b;
    x[p] -= q * x[!p];
    y[p] -= q * y[!p];
    a = b;
    b = r;
  }
  if (c % a)
    printf("-1\n");
  else
    printf("%lli %lli\n", x[!p] * c / a, y[!p] * c / a);
  return 0;
}

```