with cte_hacker_name(hacker_id, name, challenge_id, difficulty_level, score)
as
(
select
    H.hacker_id, 
    H.name, 
    C.challenge_id, 
    C.difficulty_level,
    S.score 

FROM 
    challenges AS C JOIN difficulty AS D ON C.difficulty_level = D.difficulty_level
    join submissions s on s.challenge_id = c.challenge_id
    join hackers H on H.hacker_id = C.hacker_id

where 
    d.score= s.score
)
SELECT 
    hacker_id, name
FROM 
    cte_hacker_name
GROUP BY 
    hacker_id, name
HAVING 
    COUNT(*) > 1
ORDER BY 
    COUNT(*) DESC, hacker_id ASC;