select NAME, rank, count, surname_entries.id 'wiki_id' from surnames left join surname_entries on surname_entries.surname = surnames.name and surname_entries.entry_type = 1 where index2 = '52' order by rank asc limit 0,100 Surnames > 5 > 52 > 522

Surnames > 5 > 52 > 522

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Full Surname List Starting with 522(Alphabetical)