Optically pure $(S)$-2-iodooctane is stirred in acetone with a very large excess of radioactive $\mathrm{Na}{}^{\ast }\mathrm{I}$, so the labelled iodide concentration stays essentially constant. Every substitution is a clean SN2 reaction with complete inversion, and the rate constant for substitution is the same for every molecule, whether or not it already carries a label. At a certain time $70\%$ of the substrate molecules are still unlabelled. What is the optical purity (enantiomeric excess) of the substrate then, in percent? Give the answer as an integer.

Numerical answer type, tolerance 0.5.