Bézout's identity and Extended Euclidean algorithm for polynomials

I'm trying to find the $U, V$ such that $a(x)U + m(x)V = d(x)$.

$$\begin{align*} a(x) &= x^4+1\\ m(x) &= x^3+3x+1\\ d(x) &= \gcd(a(x),m(x)) \end{align*}$$

I found $d(x)$ by using Euclidean algorithm

$$\begin{align*} x^4+1 &= (x^3+3x+1) \cdot x -(3x^2-x+1)\\ x^3+3x+1 &= (-3x^2-x+1) \cdot (\frac{-x}{3}+\frac{1}{9}) + (\frac{31x}{9}+\frac{8}{9})\\ -3x^2-x+1 &= (\frac{31x}{9}+\frac{8}{9})\cdot (\frac{-27x}{31} - \frac{63}{961})+ \frac{1017}{961}\\ \end{align*}$$

so because the remainder is equal to a constant, $d(x)= 1$.

Then I try to solve the equation $a(x)U + m(x)V = d(x)$ by using the Extended Euclidean algorithm, and more precisely by using the table (shown here: https://www.numbertheory.org/php/euclid.html)

and with all the values put in the table looks like this: https://imgur.com/a/8Wku6 (sry I don't know how to put tables into stackexchange)

I know the answer should be $(x^3+3x+1) \cdot (-\frac{31x^3}{113}+\frac{8x^2}{113}- \frac{13x}{113}+\frac{7}{113}) + (x^4+1) \cdot (\frac{31x^2}{113} - \frac{8x}{113} + \frac{106}{113}) = 1$

Where

$$U = (-\frac{31x^3}{113}+\frac{8x^2}{113}- \frac{13x}{113}+\frac{7}{113})$$
$$V = (\frac{31x^2}{113} - \frac{8x}{113} + \frac{106}{113})$$

But I don't know what I'm doing wrong with my Euclidean algorithm or Euclidean extended algorithm.

Thanks in advance