If m$($x$)=$ x$3+$ x$1+$x$0,$ J $=$ x$1+$ x$0$ and K $=$ x$4+$x$3$ what is the modular inverse of J$*$K given by Extended GCD$($ J$*$K$,$ m$($x$))$ such that $($J$*$K$)*($J$*$K$)-1=1$ mod m$($x$)$

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Provide the final value of tn$-1=($J$*$K$)-1$ and the final value sn$-1=$ m$($x$)*$m$($x$)-1*$m$($x$)=1$ mod $($J$*$K$)$

tn$-1=($J$*$K$)-1=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

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sn$-1=$ m$($x$)-1*$ m$($x$)=1$ mod $($J$*$K$)=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

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