What is the type of this surface? In my Geometric Topology lecture exams, it is asked to classify the following surface 
$$aba^{-1}cb^{-1}c^{-1}dege^{-1}fg^{-1}f^{-1}d^{-1}$$
and however, after long computation I have reduced the surface to
$$ztgt^{-1}g^{-1}z^{-1}k^{-1}r^{-1}kr$$
However, this resulting surface has to torodial component, there is the side $^{-1}$ with I couldn't get rid of and couldn't figure out why it did not cancelled out.
My question is that, is this resulting surface is anything meaningfull ? or is there any other step that I should apply to get rid of the term $z$ in this case ?
Note that, I have checked my calculations more than 5 times, and couldn't find any mistake.
Edit:
Please note that the main purğose of this question to ask how to get rid of the edge $z$ in the last expression that I got.
 A: This is very simple if you just invoke the classification of surfaces.  The key idea is that the classification says a surface is uniquely determined by its Euler characteristic and orientability, so you can pretty much identify it by just counting.
This surface is obtained by starting with a $14$-gon and then making some identifications of its vertices and edges.  The edges are identified in pairs, leaving $7$ distinct edges in the quotient.  The vertices are more complicated to count, but you can just chase through how they are identified by the edge identifications.  For instance, the starting vertex of the edge labelled $a$ must be the same as the ending vertex of the edge labelled $a^{-1}$, which is also the starting vertex of the edge labelled $c$, and so it is the same as the ending vertex of the edge labelled $c^{-1}$, and so on.  When you label all the identifications of vertices, you can find there are $4$ distinct vertices in the quotient.
So, our surface has a cell structure with $4$ vertices, $7$ edges, and $1$ face.  This means its Euler characteristic is $4-7+1=-2$, and so by the classification of closed surfaces it can only be an orientable surface of genus $2$ or a non-orientable surface of genus $3$.  To test orientability, we can just compute $H_2$: our one face has boundary $aba^{-1}cb^{-1}c^{-1}dege^{-1}fg^{-1}f^{-1}d^{-1}$, which is trivial as a chain (since then the variables can commute with each other), so the face is a cycle and $H_2$ is nontrivial.  This means our surface is orientable, so it is orientable of genus $2$.
