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

题解：初看这一题，没有想到这个是一道拓展欧几里得题，感觉答案也不唯一，仔细考虑了一下，发现了答案确实只有一个，因为他有一个确定的c，然后这个答案还与a与b的公倍数有关，cai'才知道这就是一道欧几里得的题。

思路：将ax+by+c=0，化为ax+by=-c/gcd(a,b)*gcd(a,b)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
long long exgcd(long long a, long long b, long long &x, long long &y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    long long g = exgcd(b, a % b, x, y);
    long long t;
    t = x, x = y, y = t - (a / b) * y;
    return g;
}
int main()
{
    long long a, b, c, x, y;
    cin >> a >> b >> c;
    long long t = exgcd(a, b, x, y);
    if (c % t == 0) printf("%lld %lld\n", -x * c / t, -y * c / t);
    else printf("-1\n");
    return 0;
}