Line

题目：

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

 

题意：

解二元一次方程，存在输出该解，不存在输出-1

 

思路：

将ax+by+c=0，化为ax+by=-c/gcd(a,b)*gcd(a,b)，套拓展欧几里得就行；

 

代码：

#include<stdio.h>
long long cifang(long long a,long long b,long long &x,long long &y)
{
	if(b==0)
	{
		x = 1;
		y = 0;
		return a;
	}
	long long t;
	t = cifang(b,a%b,x,y);
	long long temp = y;
	y = x-(a/b)*y;
	x  = temp;
	return t;
}
int main()
{
	long long i,j,a,b,c;
	while(~scanf("%lld%lld%lld",&a,&b,&c))
	{
		long long x,y;
		long long t =cifang(a,b,x,y);
		if(-c%t!=0)
			printf("-1\n");
		else printf("%lld %lld\n",(-c)/t*x,(-c)/t*y);
	}
	return 0;
}