Line codeforces7C

这道题相当于是一道扩展gcd的模板提。首先要知道什么是扩展gcd，按照目前的理解来说，扩展gcd是来求一组二元不定方程组的解，这个跟逆元有关，对任意的a，b和它俩的gcd(a,b),总能找到一组解使得ax+by=gcd(a,b),同时利用求gcd的方法我们知道，bx1+(a%b)y1=gcd(b,a%b),我们知道的是gcd(a,b)=gcd(b,a%b),所以将方程组化简就得到ax+by=bx1+(a-(a/b)*b)y1,然后可以得出一个结论就是x=y1,y=x-(a/b)y1。这样的话就求出了一组解使得ax+by=gcd(a,b),当然a和b是不变的，然后推广到任意一个补丁方程组ax+by+c=0都可以利用上面的那个方程组的解来得出下一个方程组的解。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    ll d;
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    d=exgcd(b,a%b,x,y);
    ll c=x;
    x=y;
    y=c-(a/b)*y;
    return d;
}
int main()
{
    ll a,b,c,d,x,y;
    scanf(\"%lld%lld%lld\",&a,&b,&c);
    d=exgcd(a,b,x,y);
    if(c%d!=0)
    {
        printf(\"-1\\n\");
    }
    else
    {
        printf(\"%lld %lld\\n\",-x*(c/d),-y*(c/d));
    }
    return 0;
}

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 5·1018inclusive, 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