7. $A_{1}(i=1,2,3,4)$ are subsets of the set $S=\{1,2, \cdots, 2005\}$, $F$ is the set of all ordered quadruples $\left(A_{1}, A_{2}, A_{3}, A_{1}\right)$, then the value of $\sum_{1}\left|A_{1} \cup A_{2} \cup A_{3} \cup A_{4}\right|$ is $\qquad$ .
7. $2^{8016} \times 2005 \times 15$ For any $k \in S=\{1,2,3, \cdots, 2005\}$, if $k \notin A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$, then $k \notin$ $A_{i}(i=1,2,3,4)$. The subsets of $S \backslash\{k\}$ have $2^{2004}$, generating $2^{2004 \times 4}$ ordered quadruples $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, the subsets of $S$ have $2^{2005}$, generating $2^{2005 \times 4}$ ordered quadruples $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, thus there are $2^{2005 \times 4}-2^{2004 \times 4}$ ordered quadruples $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$. Making $k \in A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$, each element $k$ is counted $2^{2005 \times 4}-2^{2004 \times 4}$ times, hence $\sum_{\left(A_{1}, A_{2} \cdot A_{3}, A_{4}\right) \in F} \mid A_{1} \cup A_{2} \cup A_{3} \cup A_{4}!=2005\left(2^{2005 \times 4}-2^{2004 \times 4}\right)=2^{8016} \times 2005 \times 15$.