<s>[INST] Inverse Z transform of $\frac {-81}{z^4(z-3)(z-1)}$ The question given like this:

Find the convolution of $3^nu(-n+3)*u(n-2)$

Attempt:
I tried to solve it through z transform method
Let $Y(z)=X1(z)X2(z)$
Now $Z$ transform of $x1(n)=3^nu(-n+3)$ will be $\frac{-81z^{-3}}{z-3}$
Z transform of $u(n-2)=\frac{1}{z(z-1)}$
now $Y(z)=\frac {-81}{z^4(z-3)(z-1)}$ 
I tried to take inverse z transform of this function by residue theorem like this
$Residue at z=3,will be \frac{-3^{n-1}}{2}$.
$Residue at z=1,will be \frac{-81}{2}$
$Residue at z=0,will be -40$
BUT the solution given like this
$$
y(n)=
\begin{cases}
\frac{81}{2},n>=5\\
\frac{3^{n-1}}{2},n<5\\
\end{cases}
$$
Now What mistake i am doing,any other method other than Residue throem will be helpful.
Thanks
 [/INST] $$Y(z) = \frac{-81}{z^4(z-3)(z-1)}$$
$$X(z) = z^4Y(z) =\frac{-81}{(z-3)(z-1)}$$
$$X(z) = \frac{-81}{(z-3)(z-1)} = \frac{A}{z-1} + \frac{B}{z-3} $$
$$A = \frac{81}{2}; B=\frac{-81}{2}$$
$$x(n) = \frac{81}{2}[1-3^{n-1}]u(n-1)$$
$$ X(z) = z^4Y(z)$$
$$x(n-4) = y(n)$$
$$y(n) = \frac{81}{2}[1-3^{n-5}]u(n-5)$$
</s>