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

Solution (in JAVA):

Solution:

// ~/BAU/ACM-ICPC/Teams/A++/BlackBurn95

import java.io.*;
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
                                              // (new FileReader("input.in"));
        StringBuilder out = new StringBuilder();
        StringTokenizer tk;
        
        tk = new StringTokenizer(in.readLine());
        long a = parseLong(tk.nextToken()),b = parseLong(tk.nextToken()),c = -parseLong(tk.nextToken());
        
        long d = gcd(a,b);
        if(c%d!=0) {
            System.out.println("-1");
            return;
        }
        
        long [] A = xgcd(a/d,b/d);
        long x0 = A[0],y0 = A[1];
        
        long x = c/d*x0 + b/d;
        long y = c/d*y0 - a/d;
        
        System.out.println(x+" "+y);
    }
    
    static long gcd(long a,long b) {
        return b==0 ? a : gcd(b,a%b);
    }
    
    static long[] xgcd(long a,long b) {
        long [] A = new long[3];
        
        if(b==0) {
            A[0] = 1;
            A[1] = 0;
            A[2] = a;
            return A;
        }
        
        A = xgcd(b,a%b);
        long t = A[0];
        A[0] = A[1];
        A[1]  = t-(a/b)*A[1];
        
        return A;
    }
}