#include <bits/stdc++.h>
using namespace std;
void task();
int main() {
  srand(time(0));
  task();
  return 0;
}
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
const int M = 1e5 + 10;
int a, b, c;
long long int gcd(long long int a, long long int b, long long int &x,
                  long long int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int g = gcd(b, a % b, x, y);
  long long int xx = y;
  long long int yy = x - a / b * y;
  x = xx;
  y = yy;
  return g;
}
void task() {
  scanf("%d %d %d", &a, &b, &c);
  if (c == 0) {
    printf("0 0");
    return;
  }
  long long int x, y;
  long long int g = gcd(abs(a), abs(b), x, y);
  if (abs(c) % g != 0) {
    printf("-1");
    return;
  }
  if (c > 0) x *= -1, y *= -1;
  if (a < 0) x *= -1;
  if (b < 0) y *= -1;
  long long int k = abs(c) / g;
  printf("%I64d %I64d", x * k, y * k);
}

### Prompt

Please formulate 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;
void task();
int main() {
  srand(time(0));
  task();
  return 0;
}
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
const int M = 1e5 + 10;
int a, b, c;
long long int gcd(long long int a, long long int b, long long int &x,
                  long long int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int g = gcd(b, a % b, x, y);
  long long int xx = y;
  long long int yy = x - a / b * y;
  x = xx;
  y = yy;
  return g;
}
void task() {
  scanf("%d %d %d", &a, &b, &c);
  if (c == 0) {
    printf("0 0");
    return;
  }
  long long int x, y;
  long long int g = gcd(abs(a), abs(b), x, y);
  if (abs(c) % g != 0) {
    printf("-1");
    return;
  }
  if (c > 0) x *= -1, y *= -1;
  if (a < 0) x *= -1;
  if (b < 0) y *= -1;
  long long int k = abs(c) / g;
  printf("%I64d %I64d", x * k, y * k);
}

```