Let $(R,m)$ be a noetherian local ring with the maximal ideal $m$ and $I$ an $R$-ideal such that $\sqrt{I}\neq m$. Then, is it true that $m$ contains a non-zero-divisor on $R/(I:m^{\infty})$ . The notation $I:J^{\infty}$ is the notation used to indicate the least $n$ such that $IJ^{n+1}=IJ^n$, also called the saturation of $J$ with respect to $I$.

I know that there exists an $R-$linear map from $R/I\to R/I.a^n$, for some element $a\in R$. I also know that $I:J=\cap_{i=1}^nI:a_i^{\infty}$, where the ideal $J$ is generated by $\{a_1,a_2\ldots,a_n\}$. How do we proceed using these facts? Suppose $m=(a_1,a_2,\ldots,a_n)$, then $R/(I:m^{\infty})=\cap_{i=1}^n R/(I:a_i^{\infty})$. What is the signifiance of the condition $\sqrt{I}\neq m$? Any hints? thanks beforehand.