Why it have to $n=2m$ for discrete sine transform

Question

If we have vector $x$ with $n$ elements that has DFT $y$, we can write $$y(k)=\sum\limits_{j = 0}^{n = 1} {{x_0}(j)\omega _n^{kj}}, \quad k=0,1,\dots,n-1.$$ Then, the discrete sine transform for $n=2m$ as $$y(k) = \sum\limits_{j = 1}^{m - 1} {\sin \left( {\frac{{kj\pi }}{m}} \right)x(j)}, \quad k=1,2,\dots,m-1.$$ My question is, why it choose $n=2m$ for the DST? Can you help me? Something that I know, we choose $n=2^\gamma$ for any integer $\gamma$ to solve DFT with FFT.