#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-8;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return f ? -x : x;
}
long long exgcd(int a, int b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y;
  y = z - y * (a / b);
  return d;
}
int a, b, c, d;
long long x, y;
int main() {
  a = read();
  b = read();
  c = -read();
  d = exgcd(a, b, x, y);
  if (c % d) {
    printf("-1");
  } else {
    printf("%I64d %I64d", x * (long long)c / d, y * (long long)c / d);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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>
using namespace std;
const long double eps = 1e-8;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return f ? -x : x;
}
long long exgcd(int a, int b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y;
  y = z - y * (a / b);
  return d;
}
int a, b, c, d;
long long x, y;
int main() {
  a = read();
  b = read();
  c = -read();
  d = exgcd(a, b, x, y);
  if (c % d) {
    printf("-1");
  } else {
    printf("%I64d %I64d", x * (long long)c / d, y * (long long)c / d);
  }
  return 0;
}

```