##Instruction:
Database: student_transcripts_tracking
Total Tables: 11
Total Columns: 56

Table Schemas:

student_transcripts_tracking contains tables such as Addresses, Courses, Departments, Degree_Programs, Sections, Semesters, Students, Student_Enrolment, Student_Enrolment_Courses, Transcripts, Transcript_Contents.
Table Addresses has columns such as address_id, line_1, line_2, line_3, city, zip_postcode, state_province_county, country, other_address_details.
address_id is the primary key.
Table Courses has columns such as course_id, course_name, course_description, other_details.
course_id is the primary key.
Table Departments has columns such as department_id, department_name, department_description, other_details.
department_id is the primary key.
Table Degree_Programs has columns such as degree_program_id, department_id, degree_summary_name, degree_summary_description, other_details.
degree_program_id is the primary key.
Table Sections has columns such as section_id, course_id, section_name, section_description, other_details.
section_id is the primary key.
Table Semesters has columns such as semester_id, semester_name, semester_description, other_details.
semester_id is the primary key.
Table Students has columns such as student_id, current_address_id, permanent_address_id, first_name, middle_name, last_name, cell_mobile_number, email_address, ssn, date_first_registered, date_left, other_student_details.
student_id is the primary key.
Table Student_Enrolment has columns such as student_enrolment_id, degree_program_id, semester_id, student_id, other_details.
student_enrolment_id is the primary key.
Table Student_Enrolment_Courses has columns such as student_course_id, course_id, student_enrolment_id.
student_course_id is the primary key.
Table Transcripts has columns such as transcript_id, transcript_date, other_details.
transcript_id is the primary key.
Table Transcript_Contents has columns such as student_course_id, transcript_id.
The department_id of Degree_Programs is the foreign key of department_id of Departments.
The course_id of Sections is the foreign key of course_id of Courses.
The permanent_address_id of Students is the foreign key of address_id of Addresses.
The current_address_id of Students is the foreign key of address_id of Addresses.
The student_id of Student_Enrolment is the foreign key of student_id of Students.
The semester_id of Student_Enrolment is the foreign key of semester_id of Semesters.
The degree_program_id of Student_Enrolment is the foreign key of degree_program_id of Degree_Programs.
The student_enrolment_id of Student_Enrolment_Courses is the foreign key of student_enrolment_id of Student_Enrolment.
The course_id of Student_Enrolment_Courses is the foreign key of course_id of Courses.
The transcript_id of Transcript_Contents is the foreign key of transcript_id of Transcripts.
The student_course_id of Transcript_Contents is the foreign key of student_course_id of Student_Enrolment_Courses.

##Role Access Policy (Column-Level):
Role: DataOperator_1
Accessible Columns: Addresses: address_id, city, line_1, line_2, state_province_county, zip_postcode; Courses: ALL COLUMNS; Degree_Programs: ALL COLUMNS; Departments: ALL COLUMNS; Sections: ALL COLUMNS; Semesters: other_details, semester_id; Student_Enrolment_Courses: course_id, student_course_id; Students: ALL COLUMNS; Transcript_Contents: ALL COLUMNS; Transcripts: ALL COLUMNS

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Which address holds the most number of students currently? List the address id and all lines.
Answer: SELECT T1.address_id ,  T1.line_1 ,  T1.line_2 FROM Addresses AS T1 JOIN Students AS T2 ON T1.address_id  =  T2.current_address_id GROUP BY T1.address_id ORDER BY count(*) DESC LIMIT 1
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T1.address_id ,  T1.line_1 ,  T1.line_2 FROM Addresses AS T1 JOIN Students AS T2 ON T1.address_id  =  T2.current_address_id GROUP BY T1.address_id ORDER BY count(*) DESC LIMIT 1