扩展欧几里得算法

C. Line

time limit per test

 1 second

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

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

#include<stdio.h>  #include<iostream>  #include<string>  #include<algorithm>  #include<string.h>  #include<map>   #include<math.h>  #include<set>  #include<queue>  #include<vector>  #include<stack>  #define MAX 50005#define ll long longusing namespace std;/* 对于不完全为0的非负整数a，b，gcd(a, b)表示a, b的最大 公约数，必定存在整数对x，y，满足a*x+b*y==gcd(a, b) 求解不定方程；如a*x+b*y=c; 已知a, b, c的值求x和y的值 a*x+b*y=gcd(a, b)*c/gcd(a, b); 最后转化为 a*x/(c/gcd(a, b))+b*y/(c/gcd(a, b))=gcd(a, b); 最后求出的解x0，y0乘上c/gcd(a, b)就是最终的结果了 x1=x0*c/gcd(a, b); y1=y0*c/gcd(a, b); *///求一组 x 和 y 使得：a*x + b*y = gcd(a,b)  ll egcd(ll a,ll b,ll &x,ll &y)//扩展欧几里得算法，返回的值为a,b最大公约数 {if(b==0){x=1;y=0;return a;}ll ans=egcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}int main(){ll a,b,c,x,y,t;while(cin>>a>>b>>c){t=egcd(a,b,x,y);//t=gcd(a,b);if(c%t!=0)printf("-1");elseprintf("%lld %lld\n",-x*c/t,-y*c/t);}}