For non-empty subsets $A,B \subset \mathbb{Z}$ define \[A+B=\{a+b:a\in A, b\in B\},\ A-B=\{a-b:a\in A, b\in B\}.\] In the sequel we work with non-empty finite subsets of $\mathbb{Z}$. Prove that we can cover $B$ by at most $\frac{|A+B|}{|A|}$ translates of $A-A$, i.e. there exists $X\subset Z$ with $|X|\leq \frac{|A+B|}{|A|}$ such that \[B\subseteq \cup_{x\in X} (x+(A-A))=X+A-A.\]