Distribution of $(X_1-\overline{X})/S$ where $X_i$ are i.i.d. normal

Question

I have encountered the statement says that the statistics

$${{X_1-\overline{X}} \over {S}},$$ where $X_i$ are i.i.d normal, is ancillary (where $S$ is the sample variance square root).

I know that $(X_1-\overline{X})$ is normal, but I have no idea what the distribution is when divided by $S$. It looks might be $t$ distribution, but since the nominator and denominator are not independent, this may not be true.

Does anyone have any idea? Thanks in advance!