Распаковка архива форматом ARJ

public class Stream {
    List fileList;
    private static final String INPUT_ZIP_FILE = "C:\\java_projects\\FILE.ARJ";
    private static final String OUTPUT_FOLDER = "C:\\java_projects\\UNZIPPED_file";
    public static void main( String[] args ) throws IOException, ArchiveException
    {
        Stream unArj = new Stream();
        unArj.unArjIt(INPUT_ZIP_FILE,OUTPUT_FOLDER);
    }
    
public void unArjIt(String arjFile, String outputFolder) throws ArchiveException{
     byte[] buffer = new byte[4096];
     try{
        //create output directory is not exists
        File folder = new File(outputFolder);
        if(!folder.exists()){
            folder.mkdir();
        }
        //get the zip file content
        ArjArchiveInputStream arj = new ArjArchiveInputStream(new FileInputStream(arjFile));
        ArjArchiveEntry arje = arj.getNextEntry();
        while(arje != null)
        {
            String fileName = arje.getName();
            File newFile = new File(outputFolder + File.separator + fileName);
            System.out.println("file unzip : "+ newFile.getAbsoluteFile());
            new File(newFile.getParent()).mkdirs();
            FileOutputStream fos = new FileOutputStream(newFile);
            int len;
            while((len = arj.read(buffer)) > 0)
            {
                fos.write(buffer, 0, len);
            }
            fos.close();
            arje = arj.getNextEntry();
        }
        arj.close();
        System.out.println("Done");
    }catch(IOException ex){
       ex.printStackTrace();
    }
}
}