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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
void __never(int a) { printf("\nOPS %d", a); }
long long dgcd(long long a, long long b, long long& x, long long& y) {
  long long X = 0, Y = 1;
  x = 1, y = 0;
  while (b) {
    long long q = a / b;
    a -= b * q;
    swap(a, b);
    x -= X * q;
    swap(x, X);
    y -= Y * q;
    swap(y, Y);
  }
  return a;
}
long long A, B, C;
long long sa, sb, sc;
long long sign(long long a) {
  if (a > 0) return 1;
  if (a < 0) return -1;
  return 0;
}
void init() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void sol() {
  C = -C;
  sa = sign(A);
  sb = sign(B);
  sc = sign(C);
  if (A < 0) A = -A;
  if (B < 0) B = -B;
  if (C < 0) C = -C;
  if (C == 0) {
    printf("0 0");
  } else {
    long long x = 0, y = 0;
    long long gcd = dgcd(A, B, x, y);
    if ((C % gcd) != 0)
      printf("-1");
    else {
      long long mul = C / gcd;
      x = x * sa * sc * mul;
      y = y * sb * sc * mul;
      cout << x << " " << y;
    }
  }
}
int main() {
  cin >> A >> B >> C;
  sol();
  return 0;
}