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select h.hacker_id, h.name from submissions s join challenges c on s.challenge_id = c.challenge_id join difficulty d on d.difficulty_level = c.difficulty_level join hackers h on h.hacker_id = s.hacker_id where s.score = d.score group by h.hacker_id, h.name having count(h.hacker_id) > 1 order by count(s.hacker_id) desc, s.hacker_id asc;
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