#include <bits/stdc++.h>
using namespace std;
string en = "\n";
string sp = " ";
string t = "hello";
string Y = "YES\n";
string N = "NO\n";
int dx[] = {-1, 0, 0, 1, -1, -1, 1, 1}, Row, Col;
int dy[] = {0, -1, 1, 0, -1, 1, -1, 1};
int kx[] = {-2, -2, 2, 2, -1, -1, 1, 1};
int ky[] = {-1, 1, -1, 1, -2, 2, -2, 2};
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
bool isprm(LL a) {
  for (LL i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
bool palin(string a) {
  for (int i = 0; i < a.size(); i++) {
    if (a.at(i) != a.at(a.size() - i - 1)) return 0;
  }
  return 1;
}
LL binpow(LL a, LL b) {
  LL r = 1;
  while (b > 0) {
    if (b & 1) r = r * a;
    a = a * a;
    b >>= 1;
  }
  return r;
}
LL binpow(LL a, LL b, LL m) {
  a %= m;
  LL r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return r;
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
LL xgcd(LL a, LL b, LL &x, LL &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  LL x1, y1;
  LL d = xgcd(b, a % b, x1, y1);
  x = y1, y = x1 - y1 * (a / b);
  return d;
}
int main() {
  LL i, j, k, n, m, l, s = 0, x, y, tc = 1, p, q, a, b, c = 0, d;
  bool f = 0, ff = 0;
  string st, sa, sb, sc;
  char ch;
  while (cin >> a >> b >> c) {
    d = xgcd(a, b, x, y);
    c *= -1;
    if (c % d == 0) {
      a = a / d, b = b / d, c = c / d;
      d = xgcd(a, b, x, y);
      cout << x * c << sp << y * c << en;
    } else
      cout << -1 << en;
  }
}

### 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;
string en = "\n";
string sp = " ";
string t = "hello";
string Y = "YES\n";
string N = "NO\n";
int dx[] = {-1, 0, 0, 1, -1, -1, 1, 1}, Row, Col;
int dy[] = {0, -1, 1, 0, -1, 1, -1, 1};
int kx[] = {-2, -2, 2, 2, -1, -1, 1, 1};
int ky[] = {-1, 1, -1, 1, -2, 2, -2, 2};
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
bool isprm(LL a) {
  for (LL i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
bool palin(string a) {
  for (int i = 0; i < a.size(); i++) {
    if (a.at(i) != a.at(a.size() - i - 1)) return 0;
  }
  return 1;
}
LL binpow(LL a, LL b) {
  LL r = 1;
  while (b > 0) {
    if (b & 1) r = r * a;
    a = a * a;
    b >>= 1;
  }
  return r;
}
LL binpow(LL a, LL b, LL m) {
  a %= m;
  LL r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return r;
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
LL xgcd(LL a, LL b, LL &x, LL &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  LL x1, y1;
  LL d = xgcd(b, a % b, x1, y1);
  x = y1, y = x1 - y1 * (a / b);
  return d;
}
int main() {
  LL i, j, k, n, m, l, s = 0, x, y, tc = 1, p, q, a, b, c = 0, d;
  bool f = 0, ff = 0;
  string st, sa, sb, sc;
  char ch;
  while (cin >> a >> b >> c) {
    d = xgcd(a, b, x, y);
    c *= -1;
    if (c % d == 0) {
      a = a / d, b = b / d, c = c / d;
      d = xgcd(a, b, x, y);
      cout << x * c << sp << y * c << en;
    } else
      cout << -1 << en;
  }
}

```