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>
using namespace std;
long long A, B, C, g, x, y;
long long gcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
int main() {
  scanf("%I64d%I64d%I64d", &A, &B, &C);
  if (A == 0) {
    if (C % B != 0) {
      puts("-1");
      return 0;
    }
    printf("0 %I64d\n", -C / B);
    return 0;
  }
  if (B == 0) {
    if (C % A != 0) {
      puts("-1");
      return 0;
    }
    printf("%I64d 0\n", -C / A);
    return 0;
  }
  if (A < 0 && B < 0) {
    A *= -1;
    B *= -1;
    C *= -1;
  }
  if (A < 0 && B > 0) {
    A *= -1;
    B *= -1;
    C *= -1;
  }
  g = gcd(A, abs(B), x, y);
  if (C % g != 0) {
    puts("-1");
    return 0;
  }
  if (A > 0 && B > 0) {
    printf("%I64d %I64d\n", -x * (C / g), -y * (C / g));
    return 0;
  }
  if (A > 0 && B < 0) {
    printf("%I64d %I64d\n", -x * (C / g), y * (C / g));
  }
  return 0;
}