Top 5 Customers Subquery

# This subquery shows how many of the top 5 customers are based within each country.
SELECT D.country,
COUNT(DISTINCT A.customer_id) as all_customer_count,
COUNT(DISTINCT Top_5_customers) as Top_customer_count 
FROM Customer A
INNER Join address B ON A.address_id=B.address_id
INNER JOIN city C ON B.city_id=C.city_id
INNER JOIN Country D ON C.country_id=D.country_id
LEFT JOIN
(SELECT A.customer_id,A.first_name AS customer_first_name,
A.last_name AS Customer_last_name,
D.country,C.city,SUM(E.amount) as total_amount_paid 
From Customer A
INNER JOIN address B ON A.address_id = B.address_id 
INNER JOIN city C ON B.city_id = C.city_id           
INNER JOIN country D ON C.country_ID = D. country_ID 
INNER JOIN payment E ON A.customer_id=E.customer_id 
WHERE city IN ('Aurora','Atlixco','Xintai','Adoni', 
'Dhule(Dhulia)','Kurashiki','Pingxiang',
'Sivas','Celaya','So Leopoldo')
GROUP BY city, country,customer_first_name,
customer_last_name,A.customer_id
ORDER by total_amount_paid DESC
LIMIT 5) as Top_5_Customers ON A.customer_id=top_5_customers.customer_id 
GROUP BY D.Country
HAVING COUNT (top_5_customers)>0
ORDER BY (Top_customer_count),(all_customer_count) DESC