For each employee ID, find the last name of all employees making more money than that employee. Solve it two ways: 1) using only WHERE-based join (i.e., no INNER/OUTER/CROSS JOIN) and 2) using some form of JOIN.

departments code projects -projectid deptcode Kname descriptionmanagerid startdate subdeptof workson enddate rojectidx revenue employeeid assignedtime employees employeeid firstname lastname deptcode K salary Figure 1.5: Schema for the Employees Database. departments code projects -projectid deptcode Kname descriptionmanagerid startdate subdeptof workson enddate rojectidx revenue employeeid assignedtime employees employeeid firstname lastname deptcode K salary Figure 1.5: Schema for the Employees Database