Выборка из трех таблиц с функцией count()

CREATE TABLE office(id_office INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, 
address TEXT NOT NULL, email TEXT NOT NULL);
 
CREATE TABLE manager(id_manager INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, 
FIO TEXT NOT NULL, id_office INTEGER NOT NULL, FOREIGN KEY(id_office) REFERENCES office (id_office));
 
CREATE TABLE selling(id_selling INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, 
DATE TEXT NOT NULL,amount INTEGER NOT NULL,
sum_of_sale INTEGER NOT NULL,comment TEXT ,
number_of_disk INTEGER NOT NULL, id_goods INTEGER NOT NULL,
id_manager INTEGER NOT NULL, FOREIGN KEY(id_goods) REFERENCES goods(id_goods),
FOREIGN KEY(id_manager) REFERENCES manager(id_manager));

SELECT  o.id_office AS io, o.address, 
       COUNT(selling.id_selling) 
       FROM office AS o
       JOIN selling 
       ON selling.id_manager = (SELECT manager.id_manager FROM manager WHERE
       manager.id_office=io)  GROUP BY o.address;

SELECT o.id_office, o.address, COUNT(s.id_selling)
FROM office AS o
 
JOIN manager AS m
  ON m.id_office = o.id_office
 
LEFT JOIN selling AS s
  ON s.id_manager = m.id_manager
 
GROUP BY o.id_office, o.address