electrostatics - Potential sign question

Let's consider a positive test charge $q$ and a positive source charge $q_o$

If we take direction of $\vec{r}$ from $q_o$ to $q$, then potential energy of $q_o$ due to $q$ will be:

$$P=k\dfrac{q_oq}{r}$$

However if we reverse the direction of $\vec{r}$ from $q$ to $q_o$, will there be any change in the sign of potential energy? If yes/no , why?

Edit @Utkarshfutous and @ Demosthene:

$$P=-\int_{\infty}^{r}\vec{F}.\vec{dr}=-kqq_o\int_{\infty}^{r}\dfrac{1}{r^2}dr\cos\theta$$

Now,

if $\cos\theta$ is negative when we take direction of $\vec{r}$ from $q_o$ to $q$

then $\cos\theta$ will be positive when we take direction of $\vec{r}$ from $q$ to $q_o$

Thus the sign of potential changes. Is there something I am missing?