Select record order by Birthday and assign Row Number in SQL Server

Hi! I used below script, but I want order by Id and Age.

DECLARE @InfoTable AS TABLE(Id INT IDENTITY(1,1),LastName VARCHAR(25),Name VARCHAR(25),Birthday date,Address VARCHAR(50))
    
INSERT INTO @InfoTable VALUES('Pulodov','Rustam','1987-12-22','city Dushanbe')
INSERT INTO @InfoTable VALUES('Soliev','Firuzjon','1988-06-16','city Hisor')
INSERT INTO @InfoTable VALUES('Sharipov','Sadriddin','1991-12-08','city Dushanbe')
INSERT INTO @InfoTable VALUES('Yakubi','Hasanzod','1991-05-15','city Dushanbe')
INSERT INTO @InfoTable VALUES('Sayfurov','Karomatullo','1980-05-07','city Dushanbe')
INSERT INTO @InfoTable VALUES('Asrori','Yatim','1978-02-06','city Dushanbe')

select row_number() over(order by Id) AS Id, LastName, Name, Birthday, CONVERT(VARCHAR,CONVERT(INT,DATEDIFF(DD,Birthday,GETDATE())/365.25))Age, Address from @InfoTable order by Id, CONVERT(VARCHAR,CONVERT(INT,DATEDIFF(DD,Birthday,GETDATE())/365.25))

Output result: