Computing the cycle decomposition of the composition of permutations.

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Let $\sigma,\tau\in S_5$ be given by $$ \sigma = \begin{pmatrix}1&3&5\end{pmatrix}\begin{pmatrix}2&4\end{pmatrix},\quad \tau = \begin{pmatrix}1&5 \end{pmatrix}\begin{pmatrix}2&3\end{pmatrix}. $$ I want to find the cycle decomposition of $\tau\sigma$. So I write

$$ \tau\sigma = \begin{pmatrix}1&5 \end{pmatrix}\begin{pmatrix}2&3\end{pmatrix}\begin{pmatrix}1&3&5\end{pmatrix}\begin{pmatrix}2&4\end{pmatrix}. $$ Reading from right to left and starting with the element $1$, I see that $1$ is taken to $3$ and $3$ is taken to $2$. Now, $2$ is taken to $4$, and $4$ is fixed in the other cycles, so our first cycle is $\begin{pmatrix}1&2&4\end{pmatrix}$. Since $3$ is taken to $5$ and $5$ is taken to $1$, the next cycle would be $\begin{pmatrix}3&1\end{pmatrix}$. But these cycles aren't disjoint, so I must have done something wrong. Where is my error?

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$4$ is taken to $2$ and $2$ is taken to $3$, so it's $(1243)$.

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in Simple Form :

${ \begin{bmatrix}{1} && {2} && {3} && {4} && {5} \\ {5} && {3} && {2} && {4} && {1}\end{bmatrix} }$ ${ \begin{bmatrix}{1} && {2} && {3} && {4} && {5} \\ {3} && {4} && {5} && {2} && {1}\end{bmatrix} }$

= ${ \begin{bmatrix}{1} && {2} && {3} && {4} && {5} \\ {2} && {4} && {1} && {3} && {5}\end{bmatrix} } = (1\ 2\ 4\ 3)\, (5) = (1\ 2\ 4\ 3)$