The Tables Dept1 and Dept2 Are as Follows -A Relational Algebra Operator Applied to Tables Dept1 and Dept2

The tables Dept1 and Dept2 are as follows
$Table~Dept1\\\begin{array}{|l|l|l|l|}\hline \text { Deptno } & \text { Deptname } & \text { Deptcity } & \text { Deptmgr } \\\hline 100 & \text { Logistics } & \text { Wichita } & 666 \\\hline 200 & \text { Finance } & \text { Boise } & 555 \\\hline 300 & \text { Marketing } & \text { Pierre } & 666 \\\hline\end{array}$

$Table~Dept2\\\begin{array}{|l|l|l|l|}\hline \text { Deptno } & \text { Deptname } & \text { Deptcity } & \text { Deptmgr } \\\hline 200 & \text { Finance } & \text { Boise } & 555 \\\hline 300 & \text { Marketing } & \text { Pierre } & 666 \\\hline 400 & \text { Sales } & \text { Reno } & 666 \\\hline\end{array}$
-A relational algebra operator applied to tables Dept1 and Dept2 produces the following result: $$\begin{array} { | l | l | l | l | } \hline \text { Deptno } & \text { Deptname } & \text { Deptcity } & \text { Deptmgr } \\\hline 100 & \text { Logistics } & \text { Wichita } & 666 \\\hline 200 & \text { Finance } & \text { Boise } & 555 \\\hline 300 & \text { Marketing } & \text { Pierre } & 666 \\\hline 400 & \text { Sales } & \text { Reno } & 666 \\\hline\end{array}$$ The operation is:

A) Dept1 UNION Dept2
B) Dept1 INTERSECT Dept2
C) Dept1 DIFFERENCE Dept2
D) Dept2 DIFFERENCE Dept1